Explanation:
It is given that,
Speed of the ball, v = 10 m/s
Initial position of ball above ground, h = 20 m
(a) Let H is the maximum height reached by the ball. It can be calculated using the conservation of energy as :
[tex]\dfrac{1}{2}mv^2=mgh'[/tex]
[tex]h'=\dfrac{v^2}{2g}[/tex]
[tex]h'=\dfrac{(10)^2}{2\times 9.8}[/tex]
h' = 5.1 m
The maximum height above ground,
H = 5.1 + 20
H = 25.1 meters
So, the maximum height reached by the ball is 25.1 meters.
(b) The ball's speed as it passes the window on its way down is same as the initial speed i.e. 10 m/s.
Hence, this is the required solution.
