Question

A ballplayer catches a ball 3.4s after throwing it vertically upward. with what speed did he throw it, and what height did it reach

Answer 1

From the equation of motion, we know,

$s=ut+\frac{1}{2}at^{2}$

Where s= displacement

u= initial velocity

a= gravitational force

t= time

Displacement is 0 since the ball comes back to the same point from where it was thrown.  

A = $-9.8m/s^{2}$ since the ball is thrown upwards.

Plug the known values into the equation.

=> $0=u\left ( 3.4 \right )+\frac{1}{2}\left ( -9.81 \right )(3.4^{2})$

Solving for u gives :

u= 16.67 m/ sec ....... equation (1)

At maximum height, final velocity i.e v is 0

Time take to reach the top = $\frac{3.4}{2} = 1.7 sec$

$v^{2}=u^{2}+2as$

=> $0=(16.67)^{2}+2(-9.81)(s)$

Solving for s we get

s= 14.16 m