When a projectile is fired at an angle `theta` w.r.t horizontal with velocity u, then its vertical component:

A projectile is fired at an angle theta with the horizontal with velocity u. Deduce the expression for the maximum height reached by it.

If a projectile is fired at an angle theta with the vertical with velocity u, then maximum height attained is given by:-

A projectile is fired at an angle of 30^(@) with the horizontal such that the vertical component of its initial velocity is 80m/s. Find approximatly the velocity of the projectile at time T//4 where T is time of flight.

A projectile is fired at an angle of 30^(@) to the horizontal such that the vertical component of its initial velocity is 80m//s . Its time of fight is T . Its velocity at t=T//4 has a magnitude of nearly.

A projectile is fired with a speed u at an angle theta with the horizontal. Find its speed when its direcrtion of motion makes an angle alpha with the horizontal.

For a projectile first with velocity u at an angle theta with the horizontal.

A projectile is fired at an angle theta with horizontal. (a) Show that its trajectory is a parabola (b) Prove that for a given velocity of projection the horizontal range is same for (90^(@)-theta) (c) Obtain an expression for velocity of projectile at any instant.

A projectile is projected with speed u at an angle theta with the horizontal . The average velocity of the projectile between the instants it crosses the same level is

In case of a projectile fired at an angle equally inclined to the horizontal and vertical with velocity (u). The horizontal range is:-