Question

if the speed of a projectile at maximum height is half of the initial speed of projection, horizontal range will be? ​

Answer 1

Answer: R=u²√3/g

Explanation:

H=u²sin2¤/2g.

R= u²sin2¤/g

ucos¤=u/2

cos-¹(1/2)=60°

¤=60°

R=u²sin2(60)/g

R=u²√3/g

R=0.173u²

Answer 2

At the topmost position, only horizontal velocity exists. Thus,

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

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

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

Then, the horizontal range,

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

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

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