A ball is thrown upward with an initial velocity of `80 ms^(-1)`. After how much time will it return to ground ? Draw velocity-time graph for the ball and find from the graph
(a) the maximum height attained by ball and
(b) height of the ball after `12 s`. Take `g=10 ms^(-2)`.

Explanation

Here, u=80 sm^(-1) , At the highest point `v=0, a=- 10 ,ms^(-2)`.
Let the ball reach the highest point in time 9t) As
`v=u+at, so `0=80 +(-10) t`
`t=80//10 8 s
The ball will return to ground in time `
`=2t=2 xx 8=16 s`
Velocity of the ball at differnt instants of time will be as follws:
`At t=0`, v=80 -10 xx 0 =80 ms^(-1)`
`At t=4`s, v=80 -10 xx 4 =40 ms^(-1)`
`At t=8 s`, v=80 -10 xx =8
`At t=12s`, v=80 -10 xx 12 =40 ms^(-1)`
`At t=16s`, v=80 -10 xx 16 =80 ms^(-1)
The velcityh-time graph will be a st, line `ABC as shown in fig 2 9b) .18.
. ltvegt (a) Mamimum height attained by the ball
`=area (OPB) `=1/2 xx 8 xx 80 =320 m`
(b) Height attained after `12 s`.
`=area OAB +Area BCD`
`=23 + 1/2 (12 -8) xx (-40) =320 -80 =240 m`.