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