A ball is thrown upward with an ini tial velocity of `100 "ms"^(-1)`. After how much time will it return ? Draw velocity - time graph for the ball and find from the graph (i) the maximum height attained by the ball and (ii) height of the ball after 15 s. Take `g=10 ms^(-2)` .

Explanation

Here , `u=100 ms^(-1), g=-10 ms^(-2)`
At highest point , v=0
As v=u+gt `therefore` 0=100-10 x t
`therefore` Time taken to reach highest point , `t=100/10`=10s
The ball will return to the ground at t = 20 s
Corresponding velocity- time graph of the ball is shown in
(i)Maximum height attained the ball=Area of `triangleAOB`
`=1/2xx10xx100` =500 m
(ii)Height attained after 15s=Area of `triangleAOB`+Area of `triangleBCD`
`=500+1/2(15-10)xx(-50)=500-125`=375 m