Question

A projectile is launched with initial speed v and angle ? over level ground. The projectile's maximum height H and horizontal range D are related by the equation D = 1.8H, what was the launch angle of the projectile?

Answer 1

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

$H=\frac{v^{2}Sin^{2}\theta }{2g}$         eq-1

D = horizontal range of the projectile

horizontal range of the projectile is given as

$D=\frac{v^{2}Sin{2}\theta }{g}$                     eq-2

It is also given that

D = 1.8 H

using eq-1 and eq-2

$\frac{v^{2}Sin{2}\theta }{g} = (1.8) \frac{v^{2}Sin^{2}\theta }{2g}$

$Sin{2}\theta = (1.8) \frac{Sin^{2}\theta }{2}$

$2 Sin\theta Cos\theta= (0.9) Sin^{2}\theta$

$2 Cos\theta = (0.9) Sin\theta$

tanθ = 2.22

θ = 65.75 deg