Answer:[tex]\theta =75.52^{\circ}[/tex]
Explanation:
Given
initial speed of Launch[tex](u)[/tex]=49.7 m/s
Range of Projectile =Maximum height of Projectile
Range is given by R[tex]=\frac{u^2\sin 2\theta }{g}[/tex]
Maximum height is given by [tex]H_{max}=\frac{u^2\sin ^2\theta }{2g}[/tex]
[tex]R=H_{max}[/tex]
[tex]\frac{u^2\sin 2\theta }{g}=\frac{u^2\sin ^2\theta }{2g}[/tex]
[tex]\frac{u^2\sin 2\theta }{g}-\frac{u^2\sin ^2\theta }{2g}=0[/tex]
[tex]\frac{u^2\sin \theta }{g}\cdot \left [ 2\cos \theta -\frac{1}{2}\right ]=0[/tex]
as u cannot be zero therefore
[tex]2\cos \theta -\frac{1}{2}=0[/tex]
[tex]\cos \theta =\frac{1}{4}[/tex]
[tex]\theta =75.52^{\circ}[/tex]
