Question

The speed of a projectile at its maximum height is half of its initial speed the angle of projection

Answer 1

Let initial velocity be u and angle of projection be (theta)

Since, velocity at maximum height is uCos(theta) and we are given

uCos(theta)=u/2

Therefore, Cos(theta)=½

(Theta)=π/3

Relation between range (R) and angle of projection (theta) is-

R=u²Sin(2*(theta))/g

R=u²Sin(2*π/3)/g

R=√3u²/2g

Answer 2

At maximum height, only horizontal velocity exists. Thus,

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

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

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