Question

A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Answer 1

Approximately
`122.625\; {\rm m}` (assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`, the ball was launched from ground level, and that the drag on the ball is negligible.)

Step-by-step explanation:

Let
`v_(0)` denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly
`(-v_(0))`. The velocity of the ball would be changed from
`v` to
`(-v_(0))\!` (such that
`\Delta v = (-v_(0)) - v_(0) = (-2\, v_(0))`) within
`t = 10\; {\rm s}`.

Also because the drag on the ball is negligible, the acceleration of the ball would be
`a = -g = -9.81\; {\rm m\cdot s^(-2)}`. Thus:

`\Delta v = a\, t = 10\; {\rm s} * (-9.81\; {\rm m\cdot s^(-2)}) = -98.1\; {\rm m\cdot s^(-1)}`.

Since
`\Delta v = (-2\, v_(0))`:

`-2\, v_(0) = \Delta v = -98.1\; {\rm m\cdot s^(-1)`.

`\begin{aligned}v_(0) &= \frac{-98.1\; {\rm m\cdot s^(-1)}}{-2}= 49.05\; {\rm m \cdot s^(-1)}\end{aligned}`.

The ball reaches maximum height when its velocity is
`v_(1) = 0\; {\rm m\cdot s^(-1)}`. Apply the SUVAT equation
`x = ({v_(1)}^(2) - {v_(0)}^(2)) / (2\, a)` to find the displacement
`x` between the original position (ground level, where
`v_(0) = 49.05\; {\rm m\cdot s^(-1)}`) and the max-height position of the ball (where
`v_(1) = 0\; {\rm m\cdot s^(-1)}`.)

`\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^(-1)})^(2) - (49.05\; {\rm m\cdot s^(-1)})^(2)}{2 * (-9.81\; {\rm m\cdot s^(-2)})} \\ &\approx 122.625\; {\rm m\cdot s^(-1)}\end{aligned}`.