Answer:
Maximum height reached is given as H = 5.00 m
Explanation:
As we know that if the ball is projected at some angle with the horizontal then the range of the projectile is given by the formula
[tex]R = \frac{v^2 sin2\theta}{g}[/tex]
here we have
[tex]R = 55 m[/tex]
[tex]\theta = 20^o[/tex]
v = 40 m/s
Now we have
[tex]55 = \frac{40^2 sin40}{g}[/tex]
[tex]g = 18.7 m/s^2[/tex]
now for maximum height we have
[tex]H = \frac{v^2sin^2\theta}{2g}[/tex]
[tex]H = \frac{40^2 sin^220}{2(18.7)}[/tex]
[tex]H = 5 m[/tex]