Question

The velosity at the maximum height of a projectile is half of it's initial velosity V.what will be the range on the horizontal plane?

Answer 1

initial velocity-v
velocity at max height-initial velocity×cos theta-v/2

so,cos theta-1/2
theta-60

R-u×usin2theta÷v×v×root3÷20

Answer 2

At maximum height, only horizontal velocity exists. Thus,

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

$\displaystyle\longrightarrow\sf{\cos\theta=\dfrac{1}{2}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\theta=60^o}}}$

Then, horizontal range,

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

$\displaystyle\longrightarrow\sf{R=\dfrac{u^2\sin120^o}{g}}$

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