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 `upsilon_(0)=+20 ms^(-1)`,
`a=-g=-10 ms^(-2)`,
`upsilon = 0 ms^(-1)`
If the ball rises to height y from the point of launch, then using the equation
`upsilon^(2)=upsilon_(0)^(2)+2(y-y_(0))`
we get
`0=(20)^(2)+2(-10)(y-y_(0))`
Solving, we get, `(y-y_(0))=20m`.
(b) We can solve this part of the problem in two ways.