A ball is thrown upward with an initial velocity of `100 ms^(-1)` . After how much time will it return ? Draw velocity -time graph for the ball and find the maximum height attained by the ball.
[Assume, `g = - 10 ms ^ { - 2 }`]

Explanation

Given : ` u = 100 ms^(-1) , g = - 10 ms^(-2)`
At highest point ,
v = 0
As v = u + g t
`:. 0 = 100 - 10 xx t`
`:.` Time taken to reach highest point,
`t = (100)/(10) = 10 s`
The ball will return to the ground at t = 20 s
Velocities of the ball at different instants at time will be as follows :
At `t = 0, v = 100 - 10 xx 0 = 100 ms^(-1)`
At ` t = 5 s, v = 100 - 10 xx 5 = 50 ms^(-1)`
At ` t = 10 s, v = 100 - 10 xx 10 = 0`
At ` t = 15 s, v = 100 - 10 xx 15 = - 50 ms ^(-1)`
At `t = 20 s, v = 100 - 10 xx 20 = - 100ms^(-1)`
The velocity - time graph will be as shown in the figure .
Height attained after 15 s
= Area of `Delta AOB`
+ Area of `Delta BCD`
`= 500 +(1)/(2) (15 - 10) xx (-50)`
= 500 - 125
= 375 m