In the previous section we saw a projectile thrown upwards from a height. In this section, we will see yet another type of Projectile motion.
1. Consider fig.4.34(a) below:
• A stone is thrown into the air.
♦ It is not thrown straight up.
♦ It is not thrown at any angle with the horizontal.
■ It is thrown in the exact horizontal direction from a height (h) above the ground.
2. The stop watch is turned on at the instant when the stone is thrown.
• The position of the stone at that instant is taken as the origin ‘O’.
• A horizontal line through O is taken as the x axis.
• A vertical line through O is taken as y axis.
• The velocity of the stone at O is called initial velocity of the projectile. It is denoted as $\mathbf\small{\vec v_0}$
3. Here, the direction of $\mathbf\small{\vec v_0}$ is exactly horizontal.
• So there will not be a vertical component
• So we get: $\mathbf\small{\vec v_{0x}}$ = $\mathbf\small{\vec v_0}$
4. Initially there is horizontal velocity only
• But once the stone leaves O, it will be acted upon by gravity
• So there will be two motions:
♦ The horizontal motion with a constant velocity of $\mathbf\small{\vec v_0}$
♦ The vertical motion which is under a constant acceleration of g
5. The vertical component is responsible for taking the stone ‘vertically away’ from O
• This vertical component will be affected by the acceleration due to gravity ‘g’
• This is the acceleration vector. We can denote it as (g)$\mathbf\small{\hat j}$
• As a result, magnitude of the vertical component will go on increasing
■ Note that, the initial value of the vertical component is zero. This is because, at O, the stone was given an exact horizontal velocity
• So the horizontal component will remain constant during the entire journey.
■ Note that, the air resistance can cause opposition to the projectile motion. But for our present discussion, air resistance is considered to be negligible. So we will not take it into account here.
7. The stone was thrown when the stop watch showed '0' s. What happens to the two velocities when the stop watch shows a reading of 't' seconds?
Ans: The horizontal component will remain the same because, there is no acceleration in the horizontal direction
■ The vertical component will have a larger value because there is positive acceleration (due to gravity) in the vertical direction
• We can find it's exact value at time = 't' s
• For that, we use the familiar equation: v = v0 + at
• Thus we can write: $\mathbf\small{\vec{v_y}=\vec{v_{0y}}+\vec{a_y}\,t}$
$\mathbf\small{\Rightarrow \vec{v_y}=0+(g)\hat{j}t}$
$\mathbf\small{\Rightarrow \vec{v_y}=(-gt)\hat{j}}$
• The '-' sign is given because, the travel is towards the negative side of the y axis
• So we can write:
At any time 't', after the beginning of the journey, the magnitude of the vertical component of velocity is given by Eq.4.20: $\mathbf\small{\left | \vec{v_y} \right |=gt}$
8. At time = 't' seconds:
• The magnitude of the horizontal component remains the same
• The vertical component has a higher magnitude as given by Eq.4.20 above.
■ As a result, the resultant velocity $\mathbf\small{\vec v}$ (which is the resultant of the horizontal and vertical components) will have a larger magnitude than $\mathbf\small{\vec v_0}$. This is shown in fig.4.34(b). We see the following:
• At time = 't' seconds:
♦ The stone has reached P
♦ $\mathbf\small{\vec v}$ has a larger length than $\mathbf\small{\vec v_0}$
9. We saw how the 'velocity of the stone' varies during it's travel. Next we will see how 'it's distance from O' varies
■ First we will see the horizontal travel
(i) We have seen that the horizontal velocity remains the same.
• So we can use the familiar 'equation for uniform motion': s = vt
(ii) Thus we get:
Horizontal displacement in time 't' s = $\mathbf\small{\vec{\Delta r_x}=\vec{v_0}\times t}$
(iii) That means, the magnitude of $\mathbf\small{\vec{\Delta r_x}}$ = $\mathbf\small{ | \vec{\Delta r_x} |=\left ( | \vec{v_0}| \right )t}$
(iv) This magnitude is the distance OP'. But the distance OP' is the x coordinate of P
• So we can write: At any time 't', after the beginning of the journey, the object will be at a parallel distance of '$\mathbf\small{\left ( | \vec{v_0}| \right )t}$' from the y axis
• In other words, at any time 't', after the beginning of the journey, the x coordinate of the object is given by Eq.4.21: x = $\mathbf\small{\left ( | \vec{v_0}| \right )t}$
■ Now we will see the vertical travel
(i) The vertical travel is affected by an acceleration 'g'. So we will use the familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$
• Thus we can write:
Vertical displacement in time 't' s = $\mathbf\small{0 \times t+\frac{1}{2}g t^2}$
(iv) This magnitude is the distance P'P. But the distance PP' is the y coordinate of P
• So we can write: At any time 't', after the beginning of the journey, the object will be at a parallel distance of '$\mathbf\small{\frac{-1}{2}g t^2}$' from the x axis
• In other words, at any time 't', after the beginning of the journey, the y coordinate of the object is given by Eq.4.22: y = $\mathbf\small{\frac{-1}{2}g t^2}$
• The '-' sign is given because, the travel is towards the negative side of the y axis
10. So we are now able to specify the position of a projectile at any time 't'.
• We are able to do it by using x and y coordinates.
■ If we can eliminate 't' from Eqs.4.21 and 4.22, we will get a direct relation between x and y.
Let us try:
(i) From Eq.4.21, we get: $\mathbf\small{t=\frac{x}{|\vec{v_0}|}}$
• We can use this instead of 't' in Eq.4.22.
• We get Eq.4.23: $\mathbf\small{y=\left [ \frac{-g}{2\left ( | \vec{v_0} | \right )^2} \right ]x^2}$
(ii) Consider the quantity inside the square brackets
• 'g' and '2' are constants
• Once the stone is thrown, it's initial velocity canot be changed. So $\mathbf\small{\left | \vec{v_0} \right |}$ is a constant
• So every thing inside the square brackets are constants
• Thus the final result inside those square brackets is a constant. We will denote it as 'a'
(iv) Eq.4.23 becomes: $\mathbf\small{y=ax^2}$
Where $\mathbf\small{a=\left [ \frac{-g}{2\left ( | \vec{v_0} | \right )^2} \right ]}$
(v) But $\mathbf\small{y=ax^2}$ is the equation of a parabola. So we can write:
■ The path of a projectile is a parabola
• This is shown in fig.c
11. Time required for the whole flight (Tf):
■ This is also equal to the time required to reach the ground
(i) Let us consider the vertical motion after O. We want the time 't' required for this motion.
• The vertical distance traveled in this motion is h.
(ii) We can use the familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$
• In this motion, the initial velocity is zero. It is like the stone just dropped from a height of h
• So we can put v0 = 0
• We get: $\mathbf\small{h=0 \times t+\frac{1}{2}g{t}^2}$
$\mathbf\small{\Rightarrow h=\frac{1}{2}gt^2}$
$\mathbf\small{\Rightarrow t=\sqrt{\frac{2h}{g}}}$
• So we get Eq.4.24: $\mathbf\small{T_f=\sqrt{\frac{2h}{g}}}$
12. Horizontal range of the projectile ($\mathbf\small{|\vec R|}$):
(i) For this we consider the horizontal motion
• The horizontal component of the velocity (which is a constant value) will be effective for the entire time (Tf) of the flight
(ii) So the horizontal distance = Horizontal velocity × time
= $\mathbf\small{|\vec v_{0x}|\times T_f =|\vec v_{0}|\times T_f = |\vec v_{0}|\times \sqrt{\frac{2h}{g}} }$
■ Thus we get Eq.4.25: Range of the projectile $\mathbf\small{|\vec R|}$ = $\mathbf\small{|\vec v_{0}|\times \sqrt{\frac{2h}{g}} }$
1. Consider fig.4.34(a) below:
 |
| Fig.4.34 |
♦ It is not thrown straight up.
♦ It is not thrown at any angle with the horizontal.
■ It is thrown in the exact horizontal direction from a height (h) above the ground.
2. The stop watch is turned on at the instant when the stone is thrown.
• The position of the stone at that instant is taken as the origin ‘O’.
• A horizontal line through O is taken as the x axis.
• A vertical line through O is taken as y axis.
• The velocity of the stone at O is called initial velocity of the projectile. It is denoted as $\mathbf\small{\vec v_0}$
3. Here, the direction of $\mathbf\small{\vec v_0}$ is exactly horizontal.
• So there will not be a vertical component
• So we get: $\mathbf\small{\vec v_{0x}}$ = $\mathbf\small{\vec v_0}$
4. Initially there is horizontal velocity only
• But once the stone leaves O, it will be acted upon by gravity
• So there will be two motions:
♦ The horizontal motion with a constant velocity of $\mathbf\small{\vec v_0}$
♦ The vertical motion which is under a constant acceleration of g
5. The vertical component is responsible for taking the stone ‘vertically away’ from O
• This vertical component will be affected by the acceleration due to gravity ‘g’
• This is the acceleration vector. We can denote it as (g)$\mathbf\small{\hat j}$
• As a result, magnitude of the vertical component will go on increasing
■ Note that, the initial value of the vertical component is zero. This is because, at O, the stone was given an exact horizontal velocity
6. The horizontal component is responsible for taking the stone ‘horizontally away’ from the origin
• This component is not affected by ‘g’• So the horizontal component will remain constant during the entire journey.
■ Note that, the air resistance can cause opposition to the projectile motion. But for our present discussion, air resistance is considered to be negligible. So we will not take it into account here.
7. The stone was thrown when the stop watch showed '0' s. What happens to the two velocities when the stop watch shows a reading of 't' seconds?
Ans: The horizontal component will remain the same because, there is no acceleration in the horizontal direction
■ The vertical component will have a larger value because there is positive acceleration (due to gravity) in the vertical direction
• We can find it's exact value at time = 't' s
• For that, we use the familiar equation: v = v0 + at
• Thus we can write: $\mathbf\small{\vec{v_y}=\vec{v_{0y}}+\vec{a_y}\,t}$
$\mathbf\small{\Rightarrow \vec{v_y}=0+(g)\hat{j}t}$
$\mathbf\small{\Rightarrow \vec{v_y}=(-gt)\hat{j}}$
• The '-' sign is given because, the travel is towards the negative side of the y axis
• So we can write:
At any time 't', after the beginning of the journey, the magnitude of the vertical component of velocity is given by Eq.4.20: $\mathbf\small{\left | \vec{v_y} \right |=gt}$
8. At time = 't' seconds:
• The magnitude of the horizontal component remains the same
• The vertical component has a higher magnitude as given by Eq.4.20 above.
■ As a result, the resultant velocity $\mathbf\small{\vec v}$ (which is the resultant of the horizontal and vertical components) will have a larger magnitude than $\mathbf\small{\vec v_0}$. This is shown in fig.4.34(b). We see the following:
• At time = 't' seconds:
♦ The stone has reached P
♦ $\mathbf\small{\vec v}$ has a larger length than $\mathbf\small{\vec v_0}$
9. We saw how the 'velocity of the stone' varies during it's travel. Next we will see how 'it's distance from O' varies
■ First we will see the horizontal travel
(i) We have seen that the horizontal velocity remains the same.
• So we can use the familiar 'equation for uniform motion': s = vt
(ii) Thus we get:
Horizontal displacement in time 't' s = $\mathbf\small{\vec{\Delta r_x}=\vec{v_0}\times t}$
(iii) That means, the magnitude of $\mathbf\small{\vec{\Delta r_x}}$ = $\mathbf\small{ | \vec{\Delta r_x} |=\left ( | \vec{v_0}| \right )t}$
(iv) This magnitude is the distance OP'. But the distance OP' is the x coordinate of P
• So we can write: At any time 't', after the beginning of the journey, the object will be at a parallel distance of '$\mathbf\small{\left ( | \vec{v_0}| \right )t}$' from the y axis
• In other words, at any time 't', after the beginning of the journey, the x coordinate of the object is given by Eq.4.21: x = $\mathbf\small{\left ( | \vec{v_0}| \right )t}$
■ Now we will see the vertical travel
(i) The vertical travel is affected by an acceleration 'g'. So we will use the familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$
• Thus we can write:
Vertical displacement in time 't' s = $\mathbf\small{0 \times t+\frac{1}{2}g t^2}$
(iv) This magnitude is the distance P'P. But the distance PP' is the y coordinate of P
• So we can write: At any time 't', after the beginning of the journey, the object will be at a parallel distance of '$\mathbf\small{\frac{-1}{2}g t^2}$' from the x axis
• In other words, at any time 't', after the beginning of the journey, the y coordinate of the object is given by Eq.4.22: y = $\mathbf\small{\frac{-1}{2}g t^2}$
• The '-' sign is given because, the travel is towards the negative side of the y axis
10. So we are now able to specify the position of a projectile at any time 't'.
• We are able to do it by using x and y coordinates.
■ If we can eliminate 't' from Eqs.4.21 and 4.22, we will get a direct relation between x and y.
Let us try:
(i) From Eq.4.21, we get: $\mathbf\small{t=\frac{x}{|\vec{v_0}|}}$
• We can use this instead of 't' in Eq.4.22.
• We get Eq.4.23: $\mathbf\small{y=\left [ \frac{-g}{2\left ( | \vec{v_0} | \right )^2} \right ]x^2}$
(ii) Consider the quantity inside the square brackets
• 'g' and '2' are constants
• Once the stone is thrown, it's initial velocity canot be changed. So $\mathbf\small{\left | \vec{v_0} \right |}$ is a constant
• So every thing inside the square brackets are constants
• Thus the final result inside those square brackets is a constant. We will denote it as 'a'
(iv) Eq.4.23 becomes: $\mathbf\small{y=ax^2}$
Where $\mathbf\small{a=\left [ \frac{-g}{2\left ( | \vec{v_0} | \right )^2} \right ]}$
(v) But $\mathbf\small{y=ax^2}$ is the equation of a parabola. So we can write:
■ The path of a projectile is a parabola
• This is shown in fig.c
11. Time required for the whole flight (Tf):
■ This is also equal to the time required to reach the ground
(i) Let us consider the vertical motion after O. We want the time 't' required for this motion.
• The vertical distance traveled in this motion is h.
(ii) We can use the familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$
• In this motion, the initial velocity is zero. It is like the stone just dropped from a height of h
• So we can put v0 = 0
• We get: $\mathbf\small{h=0 \times t+\frac{1}{2}g{t}^2}$
$\mathbf\small{\Rightarrow h=\frac{1}{2}gt^2}$
$\mathbf\small{\Rightarrow t=\sqrt{\frac{2h}{g}}}$
• So we get Eq.4.24: $\mathbf\small{T_f=\sqrt{\frac{2h}{g}}}$
12. Horizontal range of the projectile ($\mathbf\small{|\vec R|}$):
(i) For this we consider the horizontal motion
• The horizontal component of the velocity (which is a constant value) will be effective for the entire time (Tf) of the flight
(ii) So the horizontal distance = Horizontal velocity × time
= $\mathbf\small{|\vec v_{0x}|\times T_f =|\vec v_{0}|\times T_f = |\vec v_{0}|\times \sqrt{\frac{2h}{g}} }$
■ Thus we get Eq.4.25: Range of the projectile $\mathbf\small{|\vec R|}$ = $\mathbf\small{|\vec v_{0}|\times \sqrt{\frac{2h}{g}} }$
Now we will see some solved examples
Solved example 4.10
A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 ms-1. Neglecting air resistance, find
(a) The time taken by the stone to reach the ground,
(b) Speed with which it hits the ground
(c) Range of the stone (Take g = 9.8 ms-2)
Solution:
Part (a):
1. We can use Eq.4.24: $\mathbf\small{T_f=\sqrt{\frac{2h}{g}}}$
2. Substituting the values, we get: Tf = 10 s
Part (b):
1. Magnitude of the horizontal velocity with which the stone hits the ground = $\mathbf\small{|\vec v_x|}$ = 15 ms-1
2. Magnitude of the vertical velocity with which the stone hits the ground:
• We can use Eq.4.20: $\mathbf\small{\left | \vec{v_y} \right |=gt}$
Here t = Tf = 10 s
• Substituting the values, we get: $\mathbf\small{\left | \vec{v_y} \right |}$ = 98 ms-1.
3. Speed (magnitude of the resultant velocity) is given by $\mathbf\small{|\vec{v}|=\sqrt{|\vec{v_x}|^2+|\vec{v_y}|^2}}$
• Substituting the values, we get: Speed = 99.14 ms-1
Part (c):
1. We can use Eq.4.25: $\mathbf\small{|\vec R|}$ = $\mathbf\small{|\vec v_{0}|\times \sqrt{\frac{2h}{g}} }$
2. Substituting the values, we get: $\mathbf\small{|\vec R|}$ = $\mathbf\small{15\times \sqrt{\frac{2 \times 490}{9.8}} }$ = 150 m
• The path of the stone is shown in fig.4.35 below. It is plotted using Eq.4.23.
• We can see that, the coordinates of the point where the stone hits the ground are: (150,-490)
Solved example 4.11
An object is thrown horizontally from the top of a tower. It strikes the ground after 3 seconds at an angle of 45° with the horizontal. Find
(a) The height of the tower
(b) The speed with which the object was thrown
[g = 9.8 ms-2]
Solution:
Part (a):
Given: Tf = 3 s
1. We can use Eq.4.24: $\mathbf\small{T_f=\sqrt{\frac{2h}{g}}}$
2. Substituting the values, we get: $\mathbf\small{3=\sqrt{\frac{2h}{9.8}}}$
• So h = 44.1 m
Part (b):
Given that, the resultant velocity at the ground makes 45° with the horizontal
1. The angle 45° indicates that magnitudes of both the horizontal and vertical components are equal
• The reason can be given using the following two statements:
(i) tan 45 is always equal to 1
(ii) $\mathbf\small{\tan \theta =\frac{|\vec{v_y}|}{|\vec v_x|}=1}$ Only when $\mathbf\small{|\vec v_y| = |\vec v_x|}$
2. We obtained the height as 44.1 m
• This height was traveled vertically in 3 s
■ What would be the velocity when t = 3 s?
3. We can use Eq.4.20: $\mathbf\small{\left | \vec{v_y} \right |=gt}$
• Here t = Tf = 3 s
• So we get: $\mathbf\small{\left | \vec{v_y} \right |=9.8 \times 3 = 29.4\,ms^{-2}}$
4. From the result in (1) we get: $\mathbf\small{| \vec{v_x}|}$ = 29.4 ms-1.
■ Throughout the travel, the horizontal velocity remains the same.
• So we can write: The object was thrown horizontally with a speed of 29.4 ms-1.
Solved example 4.12
A particle is projected horizontally with a velocity of 20 ms-1. After what time will the velocity be at an angle of 45° with the horizontal? [g = 10 ms-2]
Solution:
1. Initially, the particle is projected horizontally
• So initially, it has only the horizontal component
2. But as the travel continues, there will be both horizontal and vertical components.
• They are: $\mathbf\small{\vec{v_x} \: \text{and} \: \vec{v_y}}$
• At any instant after t = 0, the velocity of the particle will be the resultant of those two components
• And that velocity will make an angle with the horizontal
3. In this problem, we are considering the instant at which the resulting velocity makes 45° with the horizontal
• 45° indicates that, $\mathbf\small{|\vec{v_x}|= |\vec{v_y}|}$
4. But $\mathbf\small{|\vec{v_x}|}$ will be always 20 ms-1.
• So we have to find the instant at which $\mathbf\small{|\vec{v_y}|}$ is also 20 ms-1
• We can use Eq.4.20: $\mathbf\small{| \vec{v_y} |=gt}$
• Substituting the values, we get: t = 20/10 = 2 s
Solved example 4.10
A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 ms-1. Neglecting air resistance, find
(a) The time taken by the stone to reach the ground,
(b) Speed with which it hits the ground
(c) Range of the stone (Take g = 9.8 ms-2)
Solution:
Part (a):
1. We can use Eq.4.24: $\mathbf\small{T_f=\sqrt{\frac{2h}{g}}}$
2. Substituting the values, we get: Tf = 10 s
Part (b):
1. Magnitude of the horizontal velocity with which the stone hits the ground = $\mathbf\small{|\vec v_x|}$ = 15 ms-1
2. Magnitude of the vertical velocity with which the stone hits the ground:
• We can use Eq.4.20: $\mathbf\small{\left | \vec{v_y} \right |=gt}$
Here t = Tf = 10 s
• Substituting the values, we get: $\mathbf\small{\left | \vec{v_y} \right |}$ = 98 ms-1.
3. Speed (magnitude of the resultant velocity) is given by $\mathbf\small{|\vec{v}|=\sqrt{|\vec{v_x}|^2+|\vec{v_y}|^2}}$
• Substituting the values, we get: Speed = 99.14 ms-1
Part (c):
1. We can use Eq.4.25: $\mathbf\small{|\vec R|}$ = $\mathbf\small{|\vec v_{0}|\times \sqrt{\frac{2h}{g}} }$
2. Substituting the values, we get: $\mathbf\small{|\vec R|}$ = $\mathbf\small{15\times \sqrt{\frac{2 \times 490}{9.8}} }$ = 150 m
• The path of the stone is shown in fig.4.35 below. It is plotted using Eq.4.23.
 |
| Fig.4.35 |
Solved example 4.11
An object is thrown horizontally from the top of a tower. It strikes the ground after 3 seconds at an angle of 45° with the horizontal. Find
(a) The height of the tower
(b) The speed with which the object was thrown
[g = 9.8 ms-2]
Solution:
Part (a):
Given: Tf = 3 s
1. We can use Eq.4.24: $\mathbf\small{T_f=\sqrt{\frac{2h}{g}}}$
2. Substituting the values, we get: $\mathbf\small{3=\sqrt{\frac{2h}{9.8}}}$
• So h = 44.1 m
Part (b):
Given that, the resultant velocity at the ground makes 45° with the horizontal
1. The angle 45° indicates that magnitudes of both the horizontal and vertical components are equal
• The reason can be given using the following two statements:
(i) tan 45 is always equal to 1
(ii) $\mathbf\small{\tan \theta =\frac{|\vec{v_y}|}{|\vec v_x|}=1}$ Only when $\mathbf\small{|\vec v_y| = |\vec v_x|}$
2. We obtained the height as 44.1 m
• This height was traveled vertically in 3 s
■ What would be the velocity when t = 3 s?
3. We can use Eq.4.20: $\mathbf\small{\left | \vec{v_y} \right |=gt}$
• Here t = Tf = 3 s
• So we get: $\mathbf\small{\left | \vec{v_y} \right |=9.8 \times 3 = 29.4\,ms^{-2}}$
4. From the result in (1) we get: $\mathbf\small{| \vec{v_x}|}$ = 29.4 ms-1.
■ Throughout the travel, the horizontal velocity remains the same.
• So we can write: The object was thrown horizontally with a speed of 29.4 ms-1.
Solved example 4.12
A particle is projected horizontally with a velocity of 20 ms-1. After what time will the velocity be at an angle of 45° with the horizontal? [g = 10 ms-2]
Solution:
1. Initially, the particle is projected horizontally
• So initially, it has only the horizontal component
2. But as the travel continues, there will be both horizontal and vertical components.
• They are: $\mathbf\small{\vec{v_x} \: \text{and} \: \vec{v_y}}$
• At any instant after t = 0, the velocity of the particle will be the resultant of those two components
• And that velocity will make an angle with the horizontal
3. In this problem, we are considering the instant at which the resulting velocity makes 45° with the horizontal
• 45° indicates that, $\mathbf\small{|\vec{v_x}|= |\vec{v_y}|}$
4. But $\mathbf\small{|\vec{v_x}|}$ will be always 20 ms-1.
• So we have to find the instant at which $\mathbf\small{|\vec{v_y}|}$ is also 20 ms-1
• We can use Eq.4.20: $\mathbf\small{| \vec{v_y} |=gt}$
• Substituting the values, we get: t = 20/10 = 2 s
In the next section, we will see circular motion.
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