Question

The speed of projectile at its maximum height is √3/2 times of its initial speed u of projection it's range on the horizontal plane is

Answer 1

uCosx = sqrt(3)/2 * u
Cosx = sqrt(3)/2
x = 30°

R = u^2 Sin(2x) / g
= u^2 Sin(60) / g
= (u^2 / g) * [sqrt(3)/2]

Range of the projectile is (u^2 / g) * [sqrt(3)/2]

Answer 2

At maximum height, only horizontal velocity exists. Thus,

$\displaystyle\longrightarrow\sf{u\cos\theta=\dfrac{u\sqrt3}{2}}$

$\displaystyle\longrightarrow\sf{\cos\theta=\dfrac{\sqrt3}{2}}$

$\displaystyle\longrightarrow\sf{\theta=30^o}$

Then, horizontal range,

$\displaystyle\longrightarrow\sf{R=\dfrac{u^2\sin(2\theta)}{g}}$

$\displaystyle\longrightarrow\sf{R=\dfrac{u^2\sin60^o}{g}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{R=\dfrac{u^2\sqrt3}{2g}}}}$