Question

A projectile is launched over a horizontal surface in such a manner that its maximum height is 3/5 of its horizontal range. Determine the launch angle.

Answer 1

The launch angle = 67.38°

Step-by-step explanation:

For projectile motion, the range (R) and maximum height (H) reached are represented by the formulas

R = (u² sin 2θ)/g

H = (u² sin² θ)/2g

H = (3/5) R

(H/R) = (3/5)

Dividing the formula for maximum height by that of the range of a projectile

(H/R) = [(sin² θ)/2] ÷ sin 2θ

Sin 2θ = 2 sinθcosθ

(H/R) = (sin² θ)/(2×2sinθcosθ)

(H/R) = (sin² θ)/(4sinθcosθ)

(H/R) = (sinθ)/(4cosθ) = (3/5)

Tanθ = (12/5)

θ = tan⁻¹ (2.4)

θ = 67.38°