Show that maximum horizontal range of a projectie is four times the corresponding maximum height?

1 Answer
Feb 1, 2018

R_max = 4H_max

Explanation:

The horizontal range is maximum for θ = 45° and it is given by
R_max = (u^2)/g........................(1)

And,

the maximum height attained is H = (u^2(sinθ)^2)/(2g)

And here θ = 45°, so

H_max = (u^2(sin45°)^2)/(2g) = u^2/(4g) ...........................(2)

From equations (1) and (2) we get :-

R_max = 4H_max

NOTE :-

1) Here θ is the angle of projection from the Horizontal.

2) u is the magnitude of the velocity with which the projectile is projected.

3) sin45°=1/(2)^(1/2)

4) g is the acceleration due to Gravity.