If `u` is the speed of a projectile fired at an angle `60^(@)` , then what is its speed at the highest point?

In case of a projectile motion, what is the angle between the velocity and acceleration at the highest point?

A stone is projected with speed of 50 ms^(-1) at an angle of 60^(@) with the horizontal. The speed of the stone at highest point of trajectory is

The range of a projectile fired at an angle of 15^@ is 50 m. If it is fired with the same speed at an angle of 45^@ its range will be

A shell is fired from a cannon with a speed of 100 m//s at an angle 60^(@) with the horizontal (positive x -direction). At the highest point of its trajectory, the shell explodes in to two equal fragments. One of the fragments moves is the speed of the other fragment at the time of explosion.

The horizontal range and the maximum height of a projectile are equal.If the speed of projection is v_(0) ,then the speed at the highest point is (2v_(0))/(sqrt(p)) The value of "p" is

A shell is fixed from a cannon with a speed of 100 m//s at an angle 60^(@) with the horizontal (positive x-direction). At the highest point of its trajectory, the shell explodes into two equal fragments. One of the fragments moves along the negative x-direction with a speed of 50 m//s . What is the speed of the other fragment at the time of explosion.

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

A projectile thrown with an initial speed u and the angle of projection 15^@ to the horizontal has a range R . If the same projectile is thrown at an angle of 45^@ to the horizontal with speed 2 u , what will be its range ?

A projectile is fired with kinetic energy 1 k J. If the range is maximum , what is its K.E. at the highest point ?