Question

A ball player catches a ball 3.2 s after throwing it vertically upward. what height did it reach?

Answer 1

At the top of the height, the velocity is zero and acceleration is negative of acceleration due to gravity ( i.e  $-9.8 m/s^2$).

The time of the ball in air is 3.2 s, so ascending time is $3.2/2=1.6 s$.

Therefore from kinematic equation,

$v = u + gt$

Substituting the values we get,

$0= u - 9.8 (1.6)\\\\u=15.7 m/s$, Here v = 0 at top.

Now from equation,

$h=ut+\frac{1}{2} gt^2$,  here h is the height .

So,

$h=(15.7 m/s) (1.6s)-\frac{1}{2} 9.8m/s^2(1.6)^2\\\\ h=25.12m-12.54m\\\\h=12.48 m$.

Thus, the ball reached at its maximum height of 12.48 m.