Question

Tge horizontal range of a projectile is 4√3 times its maximum height. Angle of projection will be

Answer 1

Answer:

R = V^2 sin (2 theta) / g      range formula

2 g H = Vy^2 = (V sin (theta))^2    height of projectile

V^2 sin (2 theta) / g =  4 * 3^1/2 * (V sin (theta))^2 / (2 g)

sin (2 theta) = 2 * 3^1/2 (sin (theta))^2

sin (2 theta) = 3.464 *(sin theta)^2

A graphical solution gives theta = .524 rad = 30 deg

sin 60 = .866

3.464 (sin 30)^2 = .866

Or sin 2 theta = 2 * 3^1/2 sin^2 theta

2 sin theta cos theta = 2 * 3^1.2 sin^2 theta

tan theta = 1 / 3^1/2 and theta = 30 deg