Question

A ball is projected horizontally from top of a tower such that it collides with wall then with ground. If after collision with wall component of velocity perpendicular to the surface is reversed in direction without change in magnitude and component of velocity parallel to the wall remains unchanged, the ball will collide with the ground at a distance of 10 ^ 4 m from the wall (H=500m; v_{0} = 20 m/s; D=100 m) Find the value of 1?(g=10 m/s^ 2 )​

Answer 1

Note: Correct question is-

A ball is projected horizontally from top of a tower such that it collides with wall then with ground. If after collision component of velocity perpendicular to the surface is reversed in direction without change in magnitude and component of velocity parallel to the surface remains unchanged. At what distance from the wall does the ball collide with the ground?

$(H=500m, v_{0}=20m/s, D=100m)$

Given data,

1. Ball is dropped from a height, $H=500m$

2. The ball is dropped at a velocity, $v_{0} =20m/s$

3. The ball collides with the wall at a distance,$D=100m$

4. Assume, $g=10m/s^{2}$

To find the distance travelled by the ball from the wall to the ground,

We know that due to gravity $g$, an object falls freely for $t$ seconds with a final velocity $v$ from a height $H$.

$H=\frac{1}{2}gt^{2}$

$500=\frac{1}{2}\times10\timest^{2}\times t^{2}$

$t^{2}=100$

$t=10s$

Now,

$speed=\frac{distance}{time}$

$\therefore distance=speed\times time=20\times10=200m$

$x=200m$

$d=x-D=200-D=200-100=100m$

Hence, the distance travelled by the ball from the wall to the ground is $100m$

Answer 2

Note: Correct question is-

A ball is projected horizontally from top of a tower such that it collides with wall then with ground. If after collision component of velocity perpendicular to the surface is reversed in direction without change in magnitude and component of velocity parallel to the surface remains unchanged. At what distance from the wall does the ball collide with the ground?

(H=500m, Vo=20m/s, D=100m)

Given data,

1. Ball is dropped from a height, H=500m
2. The ball is dropped at a velocity, Vo =20m/s
3. The ball collides with the wall at a distance,D. = 100m
4. Assume, g=10m/s^2

To find the distance travelled by the ball from the wall to the ground,

We know that due to acceleration due gravity g , an object falls freely for t seconds with a final velocity v from a height H

$H = \frac{1}{2} g {t}^{2}$

$500 = \frac{1}{2} 10 {t}^{2}$

$100 = {t}^{2}$

$t = \sqrt{100} = 10s$

Now,

$speed = \frac{dist.}{time}$

$dist. = speed \times time$

$x = 20 \times 10 = 200m$

$d = x - D = 200 - 100 = 100m$

Hence, the distance travelled by the ball from the wall to the ground is 100m