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

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 = -10 m//s^(2)`
Using Newton’s third equation of motion
`v^(2)=u^(2) +2as`
`0=u^(2)+2(-10)xx4.05`
`therefore u^(2)=81`
u = 9 m/s 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)`
`4.05=0+(1)/(2) 10t^(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=1.8 s`