A ball is thrown vertically upwards with a velcotiy of `20 ms^(-1)` from the top of a multi-storey building. The height of the point fromwher the ball is thrown if 25m from the ground. (a) How high the ball will rise ? And (b) how long will it be before the ball hits the ground ? Take. `g=10 ms^(-2)`.

Explanation

(a) Given initial velocity u=20m/s
g=`-10m//s^(2)` (upward direction is assumed as positive)
At the highest point, final velocity,v=0
Let the ball rise to height h from where it is thrown above the building Then
`v^(2)-u^(2)=2gh`
`rArr 0^(2)-20^(2)=2xx(-10)xxh`
`rArr h-(400)/(20)-20m`
(b) Net displacement of the ball s=-Height of building =-25.0m
Now, `s=ut+(1)/(2)gt^(2)`
`rArr -25=20t+(1)/(2)xx(-10)xxt^(2)`
`rArr5t^(2)-20t-25=0`
`rArr t^(2)-4t-5=0`
`(t+1)(t-5)=0`
`tcancel=-1`
`therefore t=5s`