Question

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

Answer 1

`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.