A ball is thrown vertically upwards with a velocity of `20 ms^(-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 ms^(-2)`.

Explanation

(a)Let us take the y-axis in the vertically upward direction with zero at the ground , as shown in Fig.
Now `v_0=+20ms^(-1), a=-g =-10 ms^(-2)`, v=0
If the ball rises to height y from the point of launch , 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 by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation
`y=y_0+v_0t+1/2at^2`
Now `y_0`=25m y=0m
`v_0=20ms^(-1), a=-10ms^(-2)` , t=?
`0=25+20t+(1//2)(-10)t^2`
Or , `5t^2-20t-25=0`
Solving this quadratic equation for t , we get t=5s .