Question

A projectile is fired with an initial speed of 38.8 m/s. Find the angle of projection such that the maximum height of the projectile is equal to its horizontal range.

Answer 1

sorry...initial speed is 38.8

Answer 2

Uy=USinX
Ux=UCosX (X is the angle of projection)

Flight time, T=2t (T=total flight time,t is time taken to maximum=time taken from maximum to ground)
V=Uy+gt
0=USinx+gt
2t=2USinx/g
T=2USinx/g

Maximum Height , V=0,U=38.8m/s

V=0
V^2=Uy^2+2gS
0=Uy^2+2gS
S=(U^2)(sinx^2)/2g


Range,

R=(Vx)(T)
=(UCosx)(2UsinX/g)
=(U^2)(Sin2x)/g





Maximum height = (u^2)(sinx^2)/2g
Range = (u^2)(sin2x)/g

Take Maximum height = Range ( u=38.8m/s, g = 9.8m/s)

you can find x the angle of projection

Work it out yourself , I hope this help.