In the previous section we saw some properties of Projectile motion. In this section, we will see some solved examples. Solved example 4.9 further below shows the calculations when a projectile is thrown upwards from a height.
Solved example 4.6
Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”. Prove this statement.
Solution:
• Here, 'elevation' indicates 'initial angle of projection' θ0.
■ Case 1: Let the angle of projection exceed 45° by x°.
Then mathematically, the angle of projection θ0 = (45+x)o
■ Case 2: Let the angle of projection fall short of 45° by (the equal amount) x°
Then mathematically, the angle of projection θ0 = (45-x)o
We will consider each case separately. The steps are given below:
Case 1:
1. We can use Eq.4.19: Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
2. Substituting the angle, we get: $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2(45+x) \right )}{g}=\frac{\left ( \left | \vec{v_0} \right |^2 \sin (90+2x) \right )}{g}}$
3. But from math classes, we know that sin (90+x) = cos x
So sin (90+2x) = cos 2x.
4. So we get the range as: $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \cos 2x \right )}{g}}$
Case 2:
1. We can use Eq.4.19 again: Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
2. Substituting the angle, we get: $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2(45-x) \right )}{g}=\frac{\left ( \left | \vec{v_0} \right |^2 \sin (90-2x) \right )}{g}}$
3. But from math classes, we know that sin (90-x) = cos x
So sin (90-2x) = cos 2x.
4. So we get the range as: $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \cos 2x \right )}{g}}$
• This is same as the result in case 1
■ So the two ranges are equal.
Solved example 4.7
A projectile is fired with a speed of 'k' ms-1. The angle of projection is 57o. The range obtained is 'R'. Determine the other angle at which the the projectile should be fired with the same speed 'k' to obtain the same range 'R'
Solution:
1. The given angle is 57o. it exceeds 45 by (57-45) = 12o
2. The other angle must fall short of 45 by the same amount 12
3. So the other angle = (45-12) = 33o.
Solved example 4.8
Two objects are projected at angles 45o and 60o. The maximum heights reached are the same. What is the ratio of their initial velocities?
Solution:
1. Let
• Magnitude of the initial velocity of object 1 be $\mathbf\small{\left | \vec{{v}_{01}} \right |}$
• Magnitude of the initial velocity of object 2 be $\mathbf\small{\left | \vec{{v}_{02}} \right |}$
2. Time to reach maximum heights:
• We can use Eq.4.16: $\mathbf\small{t_m=\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$
(i) For object 1, we get: $\mathbf\small{t_{m1}=\frac{\left | \vec{v_{01}} \right |\sin 45}{g}}$
$\mathbf\small{\Rightarrow t_{m1}=\frac{\left | \vec{v_{01}} \right |}{\sqrt{2}\,g}}$
(ii) For object 2, we get: $\mathbf\small{t_{m2}=\frac{\left | \vec{v_{02}} \right |\sin 60}{g}}$
$\mathbf\small{\Rightarrow t_{m2}=\frac{\sqrt{3}\,\left | \vec{v_{02}} \right |}{{2}\,g}}$
3. But given that tm1 = tm2.
• So we can equate 2(i) and 2 (ii). We get:
$\mathbf\small{\frac{\left | \vec{v_{01}} \right |}{\sqrt{2}\,g}}$ = $\mathbf\small{\frac{\sqrt{3}\,\left | \vec{v_{02}} \right |}{{2}\,g}}$
$\mathbf\small{\Rightarrow \frac{\left | \vec{v_{01}} \right |}{\left | \vec{v_{02}} \right |}=\frac{\sqrt 3}{\sqrt 2}}$
Solved example 4.9
An object is thrown from the top of a building 10 m high. It is thrown upwards with a speed of 25 ms-1 at an angle of 40o with the horizontal.
(a) After what time will it reach the ground?
(b) What is the distance between the foot of the cliff and the point of impact on the ground?
[g = 9.81 ms-2]
Solution:
Part (a):
1. As usual, we choose the point of projection 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.
This is shown in fig.4.33 below:
2. We get the same equation Eq.4.15 for the path of the projectile:
Eq.4.15: $\mathbf\small{y=\left [ \tan\theta_0 \right ]x-\left [ \frac{g}{2\left ( \left | \vec{v_0} \right |\cos \theta_0 \right )^2} \right ]x^2}$
3. But this time, the path continues to a point Q below the x axis
• This point is at a vertical distance of 10 m below the x axis. This is because, the height of the building is 10 m
4. After being thrown from O, the object meets the x axis again at P
• We can find the time required to reach P from O.
• We can use Eq.4.18: $\mathbf\small{T_f=\frac{2\left | \vec{v_0} \right |\sin \theta_0}{g}}$
• Substituting the values, we get: Tf = 3.276 s
5. After passing P, the stone continues the flight for some more time. In the end, it falls back to the ground.
• Let us consider the vertical motion after P. We want the time 't' required for this motion.
• The vertical distance traveled in this motion is 10 m.
■ We can use the familiar equation: $\mathbf\small{s=\left | \vec{v_{0y}} \right |t+\frac{1}{2}\left | \vec{a_y}\right |t^2}$
• Here, s = 10 m.
• $\mathbf\small{|\vec{v_{0y}}|}$ = the magnitude of the vertical velocity at P
= magnitude of the vertical velocity at O = 25 sin θ0 = 16.07 ms-1
• $\mathbf\small{|\vec{a_{0y}}|}$ = g = 9.81 ms-2.
6. Substituting the values, we get: 10 = 16.07t + 0.5 × 9.81t2
⇒ 4.905t2 + 16.07t - 10 = 0
• Solving this quadratic equation, we get: t = 0.535 s or -3.811 s
• But negative time is not acceptable. So we take t = 0.535 s
7. That is., after passing P, the object travels for 0.535 s more
• So total time of travel = 3.276 + 0.535 = 3.811 s
Part (b):
1. We can find the distance OP using Eq.4.19:
Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
• Substituting the values, we get:
OP = 62.74 m
2. During the last 0.535 s, the object travels horizontally also. The distance covered during this time =
v0cos θ0 × t = 25 × cos 40 × 0.535 = 10.245 m
3. So total horizontal distance from O to Q = 62.74 + 10.245 = 72.99 m
• We can see that, the coordinates of Q are (72.99, -10)
Solved example 4.6
Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”. Prove this statement.
Solution:
• Here, 'elevation' indicates 'initial angle of projection' θ0.
■ Case 1: Let the angle of projection exceed 45° by x°.
Then mathematically, the angle of projection θ0 = (45+x)o
■ Case 2: Let the angle of projection fall short of 45° by (the equal amount) x°
Then mathematically, the angle of projection θ0 = (45-x)o
We will consider each case separately. The steps are given below:
Case 1:
1. We can use Eq.4.19: Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
2. Substituting the angle, we get: $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2(45+x) \right )}{g}=\frac{\left ( \left | \vec{v_0} \right |^2 \sin (90+2x) \right )}{g}}$
3. But from math classes, we know that sin (90+x) = cos x
So sin (90+2x) = cos 2x.
4. So we get the range as: $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \cos 2x \right )}{g}}$
Case 2:
1. We can use Eq.4.19 again: Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
2. Substituting the angle, we get: $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2(45-x) \right )}{g}=\frac{\left ( \left | \vec{v_0} \right |^2 \sin (90-2x) \right )}{g}}$
3. But from math classes, we know that sin (90-x) = cos x
So sin (90-2x) = cos 2x.
4. So we get the range as: $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \cos 2x \right )}{g}}$
• This is same as the result in case 1
■ So the two ranges are equal.
Solved example 4.7
A projectile is fired with a speed of 'k' ms-1. The angle of projection is 57o. The range obtained is 'R'. Determine the other angle at which the the projectile should be fired with the same speed 'k' to obtain the same range 'R'
Solution:
1. The given angle is 57o. it exceeds 45 by (57-45) = 12o
2. The other angle must fall short of 45 by the same amount 12
3. So the other angle = (45-12) = 33o.
Solved example 4.8
Two objects are projected at angles 45o and 60o. The maximum heights reached are the same. What is the ratio of their initial velocities?
Solution:
1. Let
• Magnitude of the initial velocity of object 1 be $\mathbf\small{\left | \vec{{v}_{01}} \right |}$
• Magnitude of the initial velocity of object 2 be $\mathbf\small{\left | \vec{{v}_{02}} \right |}$
2. Time to reach maximum heights:
• We can use Eq.4.16: $\mathbf\small{t_m=\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$
(i) For object 1, we get: $\mathbf\small{t_{m1}=\frac{\left | \vec{v_{01}} \right |\sin 45}{g}}$
$\mathbf\small{\Rightarrow t_{m1}=\frac{\left | \vec{v_{01}} \right |}{\sqrt{2}\,g}}$
(ii) For object 2, we get: $\mathbf\small{t_{m2}=\frac{\left | \vec{v_{02}} \right |\sin 60}{g}}$
$\mathbf\small{\Rightarrow t_{m2}=\frac{\sqrt{3}\,\left | \vec{v_{02}} \right |}{{2}\,g}}$
3. But given that tm1 = tm2.
• So we can equate 2(i) and 2 (ii). We get:
$\mathbf\small{\frac{\left | \vec{v_{01}} \right |}{\sqrt{2}\,g}}$ = $\mathbf\small{\frac{\sqrt{3}\,\left | \vec{v_{02}} \right |}{{2}\,g}}$
$\mathbf\small{\Rightarrow \frac{\left | \vec{v_{01}} \right |}{\left | \vec{v_{02}} \right |}=\frac{\sqrt 3}{\sqrt 2}}$
Solved example 4.9
An object is thrown from the top of a building 10 m high. It is thrown upwards with a speed of 25 ms-1 at an angle of 40o with the horizontal.
(a) After what time will it reach the ground?
(b) What is the distance between the foot of the cliff and the point of impact on the ground?
[g = 9.81 ms-2]
Solution:
Part (a):
1. As usual, we choose the point of projection 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.
This is shown in fig.4.33 below:
 |
| Fig.4.33 |
Eq.4.15: $\mathbf\small{y=\left [ \tan\theta_0 \right ]x-\left [ \frac{g}{2\left ( \left | \vec{v_0} \right |\cos \theta_0 \right )^2} \right ]x^2}$
3. But this time, the path continues to a point Q below the x axis
• This point is at a vertical distance of 10 m below the x axis. This is because, the height of the building is 10 m
4. After being thrown from O, the object meets the x axis again at P
• We can find the time required to reach P from O.
• We can use Eq.4.18: $\mathbf\small{T_f=\frac{2\left | \vec{v_0} \right |\sin \theta_0}{g}}$
• Substituting the values, we get: Tf = 3.276 s
5. After passing P, the stone continues the flight for some more time. In the end, it falls back to the ground.
• Let us consider the vertical motion after P. We want the time 't' required for this motion.
• The vertical distance traveled in this motion is 10 m.
■ We can use the familiar equation: $\mathbf\small{s=\left | \vec{v_{0y}} \right |t+\frac{1}{2}\left | \vec{a_y}\right |t^2}$
• Here, s = 10 m.
• $\mathbf\small{|\vec{v_{0y}}|}$ = the magnitude of the vertical velocity at P
= magnitude of the vertical velocity at O = 25 sin θ0 = 16.07 ms-1
• $\mathbf\small{|\vec{a_{0y}}|}$ = g = 9.81 ms-2.
6. Substituting the values, we get: 10 = 16.07t + 0.5 × 9.81t2
⇒ 4.905t2 + 16.07t - 10 = 0
• Solving this quadratic equation, we get: t = 0.535 s or -3.811 s
• But negative time is not acceptable. So we take t = 0.535 s
7. That is., after passing P, the object travels for 0.535 s more
• So total time of travel = 3.276 + 0.535 = 3.811 s
Part (b):
1. We can find the distance OP using Eq.4.19:
Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
• Substituting the values, we get:
OP = 62.74 m
2. During the last 0.535 s, the object travels horizontally also. The distance covered during this time =
v0cos θ0 × t = 25 × cos 40 × 0.535 = 10.245 m
3. So total horizontal distance from O to Q = 62.74 + 10.245 = 72.99 m
• We can see that, the coordinates of Q are (72.99, -10)
In the next section, we will see another type of projectile motion.
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