Question

A ball kept on a wall is pushed horizontally with
certain velocity and allowed to move under gravity. Sucha
motion is two-dimensional motion with ball covering
displacements in both x and y-directions. But both the x & y
direction motions are independent of each other. Hence,motion in x-direction can be considered as similar to straight
line motion with no acceleration. Motion in y-direction can be
considered as free fall under gravity.
In the figure shown, ball is pushed horizontally from a
height of 19.If there is no energy loss when particle collides at point B on ground, find the maximum
height attained by the particle after it rebounds

Answer 1

A ball kept on a wall is pushed horizontally with certain velocity and allowed to move under gravity. Such a motion is two-dimensional motion with ball covering displacements in both x and y-directions. But both the x & y direction motions are independent of each other. Hence, motion in x-direction can be considered as similar to straight line motion with no acceleration. Motion in y-direction can be considered as free fall under gravity. In the figure shown, ball is pushed horizontally from a height of 19 m. If there is no energy loss when particle collides at point B on ground, find the maximum height attained by the particle after it rebounds.

Solution

When a ball is pushed horizontally with certain velocity $v_{x}$ from height $h = 19 \, m$ its energy consists of kinetic energy and potential energy:

$$E = \frac{m v_{x}^{2}}{2} + m g h.$$

If there is no energy loss when particle collides at point B on ground energy conserves, so

$$E = \frac{m v^{2}}{2} + m g h_{\max},$$

when a ball at its maximum height.

But at maximum height ball have vertical component of velocity $v_{y} = 0$ and its horizontal component of velocity $v_{x}$ doesn't change. That's why $v = v_{x}$. So

$$\frac{m v_{x}^{2}}{2} + m g h = \frac{m v^{2}}{2} + m g h_{\max}, \quad v = v_{x}.$$ $$m g h = m g h_{\max} \rightarrow h_{\max} = h = 19 \, m.$$

Answer: 19 m.