Question

A projectile thrown from the ground has an initial speed u and its direction makes an angle θ with the horizontal. If at maximum height from the ground, the speed of the projectile is half its initial speed of projection, then the maximum height reached by the projectile is ______.

`["g = acceleration due to gravity", sin30^circ = cos60^circ = 0.5, cos30^circ = sin60^circ = sqrt3"/"2]`

Options

• `(2u^2)/g`

• `(3u^2)/(8g)`

• `u^2/g`

• `u^4/(2g)`

Answer 1

A projectile thrown from the ground has an initial speed u and its direction makes an angle θ with the horizontal. If at maximum height from the ground, the speed of the projectile is half its initial speed of projection, then the maximum height reached by the projectile is `underlinebb((3u^2)/(8g))`.

Explanation:

Given, at maximum height,

`u costheta = 1/2 u`

⇒ `costheta = 1/2 or theta = 60^circ`

∴ Maximum height, H_max = `(u^2sin^2theta)/(2g)`

= `(u^2 xx sin^2 60^circ)/(2g) = (3u^2)/(8g)` `[∵ sin60^circ = sqrt3/2]`