The trajectory of a projectile fired horizontally with velocity u is parabola given by-

Trajectory Of Horizontal Projectile

At the top of the trajectory of a projectile, the directions of its velocity and acceleration are

At the top of the trajectory of projectile the

a projectile is fired from the surface of the earth with a velocity of 5ms^(-1) and angle theta with the horizontal. Another projectile fired from another planet with a velocity of 3ms^(-1) at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth.The value of the acceleration due to gravity on the planet is in ms^(-2) is given (g=9.8 ms^(-2))

A projectile is projected from horizontal with velocity (u) making an angle 45^(@) with the horizontal direction. Find the distance of the highest point of the projectile from its starting point.

The angle of which the velocity vector of a projectile thrown with a velocity u at an angle theta to the horizontal will take with the horizontal after time t of its being thrown up is

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

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

Why does a projectile fired along the horizontal not follow a straight line path ?

The velocity at the maximum height of a projectile is half of its velocity of projection u . Its range on the horizontal plane is