Question

A projectile is thrown with velocity v at an angle theta with horizontal. When the projectile is at a height equal to the half of the maximum height,thr vertical component of the velocity of projectile is..

Answer 1

The angle at which the projectile is thrown is theta and let us denote it for this question by x.

Initial vertical velocity component u = v sinx

At the top, velocity of projectile is given by

(vsinx )2 = 2 * g * h

h = (v sinx )2/ 2 * g

On further solving and calculating the vertical velocity component at this point we have,

Velocity at top = u sinx / 21/2