Answer:
Explanation:
Given
Initial launch angle [tex]\theta =45^{\circ}[/tex]
Suppose u is the initial launching velocity
Range of Projectile is given by
[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
thus value of [tex]u^2=\frac{Rg}{\sin 2\theta }[/tex]
If this ball is thrown vertically upward with u
then maximum height reached is
[tex]h=\frac{u^2}{2g}[/tex]
[tex]h=\frac{\frac{Rg}{\sin 2\theta }}{2g}[/tex]
[tex]h=\frac{R}{2\sin 2\theta }[/tex]