The range of a target is found to be 20km. A shell leaves a gun with a velocity of 500m/s. What must be the angle of elevation of the gun. If the ground is level

Answer 1

I need the solution

Answer 2

Recall that the range
R = v^2/g sin2θ
so you just need to find θ such that
500^2/9.81 sin2θ = 20000

Answer 3

The range of a projectile launched at velocity V and an angle θ to the horizontal is:

R = (V^2 Sin2θ)/g

Here, R = 20km = 20,000m and V = 500m/s, so

20,000m x 9.8m/s^2 = 500^2 (m/s)^2 • Sin 2θ

So, Sin2θ = 0.784

θ = 0.4585 radians = 25.812°

The angle of the gun is 25.812 degrees to the horizontal.