Question

the speed at the maximum height of projectile is half of it's initial speed u. it's range on the horizontal plane is

Answer 1

its initial speed = u

speed at maximum heeight = vcosx = 1/2v

hence cosx = 1/2

x = 60

range = u2sin2x/g

= u2 sin(120)/10

=u2 31/2/20

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,

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

$\displaystyle\longrightarrow\sf{R=\dfrac{u^2\sin120^o}{10}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{R=\dfrac{u^2\sqrt3}{20}}}}$