# PhysicsTeacher.in

## Problem Statement

A baseball is popped straight up into the air and has a hang-time of 6 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the max height is one-half the total hang-time.)

## Solution

Hang time = Total time required for upward plus downward movement.

We get the time for upward movement as = 6/2 = 3 secs

This is the time required to reach the max height when the ball is thrown vertically upwards.
As the initial upward throwing velocity is not provided we can’t directly find out the max height.

So we will try with the downward movement.

While falling downwards: U = initial velocity = 0

And time for downward movement = time for upward movement = 3 secs

So the velocity of the ball just before touching the ground during the downward return becomes = final velocity for downward movement =
V = U + gt = 0 + 10. 3 = 30 m/s

g = 10 m/s^2

We know that final velocity of the vertical downward movement = initial throwing velocity of the concerned vertical upward movement

So we here get the initial upward throwing velocity as 30 m/s

Coming back to upward movement:

Initial upward throwing velocity = 30 m/s (as found out above)

So now we can easily find out the max height reached during upward movement
= H = U2/(2g) = 302 / (2 * 10) = 45 m

Question Set – Physics Problems for grade 9

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