A tennis ball is thrown up and reaches a height of 4.05 m before coming down. What was its initial velocity? How much total time will it take to come down? Assume `g=10m//s^(2)`

Explanation

Given
For the upward motion of the ball, the final velocity of the ball = v = 0 Distance travelled by the ball = 4.05 m
acceleration (a) `=-g=-10m//s^(2)`
Using Newton.s third equation of motion,
`v^(2)=u^(2)+2a s`
`0+u^(2)+2(-10)xx4.05`
`thereforeu^(2)=81`
The initial velocity of the ball is 9 m/s
Now let us consider the downward motion of the ball. Suppose the ball takes t seconds to come down. Now the initial velocity of the ball is zero, u = 0. Distance travelled by the ball on reaching the ground = 4.05 m . As the velocity and acceleration are in the same direction,
a = g = 10 m/s
According to Newton.s second equation of motion,
`s=ut+(1)/(2)at^(2)`
`therefore4.05=0+(1)/(2)xx10xxt^(2)`
`t^(2)=(4.05)/(5)=0.81,`
t = 0.9 s
The ball will take 0.9 s to reach the ground. It will take the same time to go up. Thus, the total time taken `2xx0.9=1.8s`