Explanation:
It is given that,
Speed of the ball, v = 10 m/s
Initial position of ball above ground, h = 20 m
We need to find the ball's maximum height above the ground. 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. Hence, this is the required solution.
