The speed of a projectile at its highest point is `v_1` and at the point half the maximum height is `v_2`. If `v_1/v_2 = sqrt(2/5)`, then find the angle of projection.

The resultant velocity of a projectile at its highest point is sqrt(6/7) times that at half the maximum height. Show that the angle of projection with respect to the horizontal is 30^@ .

The speed of a projectile when it is at its greatest height is sqrt(2//5) times its speed at half the maximum height. The angle of projection is

The speed of a projectile when it is at its greatest height is sqrt(2//5) times its speed at half the maximum height. The angle of projection is

The speed of a projectile when it is at its greatest height is sqrt(2/5) times its speed at half the maximum height. What is its angle of projection?

The velocity of a projectile when at its 2 greatest height is sqrt(2/5) of its velocity when at half of its greatest height find the angle of projection

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

[" 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 "],[" Answer: "]

A body is projected with a speed v at an angle theta with horizontal to have maximum range. What is the velocity at the highest’ point?