(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.
