Answer:
H = 8.45 m
Explanation:
Given that,
Angle of projection, [tex]\theta=37^{\circ}[/tex]
The initial velocity of the ball, u = 20 m/s
We need to find the maximum height reached by the ball. The formula for the maximum height is given by :
[tex]H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(20)^2\times \sin^2(37)}{2\times 9.8}\\\\H=8.45\ m[/tex]
So, the maximum height reached by the ball is 8.45 m.
