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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)`.

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(a) Let us take the y-axis in the vertically upward direction with zero at the ground, as shown in Fig.3.13
Now `v_(0) = + 20 ms^(-1)`
`a = -g = - 10 ms^(-2)`
`v = 0 ms^(-1)`
If the ball rises to height y from the point of launch, then using the equation
`v^(2) = v_(0)^(2) + 2a (y - y_(0))`
we get `0 = (20)^(2) + 2(-10) (y - y_(0))`
Solving, we get, `(y - y_(0)) = 20 m`.
(b) We can solve this part of the problem in two ways.
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