Question

Find the angle of projection at which the horizontal range is twice the maximum height of a projectile . . .
(a)53.4°
(b)63.4°
(c)73.4°
(d)83.4°

Answer 1

The horizontal range of a projectile is given by the equation:

R = (v^2 sin(2θ)) / g

The maximum height reached by the projectile is given by the equation:

H = (v^2 sin^2(θ)) / (2g)

Given that the horizontal range is twice the maximum height, we can write:

2H = R

Substitute the values of R and H:

2((v^2 sin^2(θ)) / (2g)) = ((v^2 sin(2θ)) / g)

Simplify this equation:

sin^2(θ) = sin(2θ)

sin^2(θ) = 2sin(θ)cos(θ)

sin(θ) = 2cos(θ)

tan(θ) = 2

θ = tan^(-1)(2)

θ ≈ 63.4°

Therefore, the angle of projection at which the horizontal range is twice the maximum height of a projectile is approximately 63.4°.

Therefore, the correct answer is (b) 63.4°.