A ball is thrown vertically upwards with a velocity of `20 m s^(-1)` from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how long will it be before the ball hits the ground? Take `g=10 m s^(-2)`

Explanation

(a)Let us take they y-axis in the vertically upward direction with zero at the ground ,as shown in fig.
Now `v_(0)=+20ms^(-1),a=-g=-10ms^(-2)`,V=0
IF the ball rises to height y from the point of lauch.then using the equation
`V^(2)=V_(0)^(2)+2as " "0=(20)^(2)+2(-10)y`
On solving ,we get ,y=20 m.
(b)The total time taken can also be calculated b y noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation
`y=y_(0)+v_(0)t+(1)/(2)at^(2)` Now ,`y_(0)`=25m y=0m
`v_(0)=20m^(-1),=-10 ms^(-2),t=?`
`v_(0)=20 ms^(-1),a=-10ms^(-2),t=?`
Or ,`5t^(2)-20t-25=0`
Solving this quadratic equation for t,we get t=5s