The horizontal range of a projectile is `2 sqrt(3)` times its maximum height. Find the angle of projection.

The horizontal range of a projectile is 4 sqrt(3) times its maximum height. Its angle of projection will be

The horizontal range is four times the maximum height attained by a projectile. The angle of projection is

A body is projected from ground with velocity u making an angle theta with horizontal. Horizontal range of body is four time to maximum height attained by body. Find out (i) angle of projection (ii) Range and maximum height in terms of u (iii) ratio of kinetic energy to potential energy of body at maximum height

The horizontal range is equal to two times maximum height of a projectile. The angle of projection ofthe projectile is:

The maximum range of projectile is 2/sqrt3 times actual range. What isthe angle of projection for the actual range ?

Assertion: The maximum horizontal range of projectile is proportional to square of velocity. Reason: The maximum horizontal range of projectile is equal to maximum height attained by projectile.

The range of a projectile for a given initial velocity is maximum when the angle of projection is 45^(@) . The range will be minimum, if the angle of projection is

The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to below in the direction of motion of the projectile, giving it a constant horizontal acceleration =g//2 . Under the same conditions of projection. Find the horizontal range of the projectile.