Answer:
65.75 deg
Explanation:
v = initial speed of launch of projectile
θ = initial angle of launch
H = maximum height of the projectile
maximum height of the projectile is given as
[tex]H=\frac{v^{2}Sin^{2}\theta }{2g}[/tex] eq-1
D = horizontal range of the projectile
horizontal range of the projectile is given as
[tex]D=\frac{v^{2}Sin{2}\theta }{g}[/tex] eq-2
It is also given that
D = 1.8 H
using eq-1 and eq-2
[tex]\frac{v^{2}Sin{2}\theta }{g} = (1.8) \frac{v^{2}Sin^{2}\theta }{2g}[/tex]
[tex]Sin{2}\theta = (1.8) \frac{Sin^{2}\theta }{2}[/tex]
[tex]2 Sin\theta Cos\theta= (0.9) Sin^{2}\theta[/tex]
[tex]2 Cos\theta = (0.9) Sin\theta[/tex]
tanθ = 2.22
θ = 65.75 deg
