A ball is thrown upward with an initial velocity of `100ms^(-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=`-10 ms^(-2)`
At highest point ,v=0
As v=u+gt `therefore=100-10xxt`
`therefore` Time taken to reah 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 `DeltaAOB=(1)/(2)xx10xx10=500 m`
(ii)HEight attained after 15s=
Area of `DeltaAOB`+Area of `DeltaBCD`
`500+(1)/(2)(15-10)xx(-50)=500-125=375m`