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

65.75 deg

Step-by-step 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=(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) (v^(2)Sin^(2)\theta )/(2g)`

`Sin{2}\theta = (1.8) (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