Question

the velocity at maximum height of a projectile is half of its initial velocity of projectile u . find its horizontal range

Answer 1

first we need to find the angle.

we are given that velocity at maximum is half of initial velocity .

we know that at maximum height, the vertical component of velocity is zero that's why it is on maximum height.

so only horizontal component remains.

horizontal component is constant in whole motion. that is ucos theta.

so ucos theta=u/2.
=>cos theta=1/2
=>theta=π/6

now range of projectile=(u^2 sin2π/6)/g.
=u^2/2g