If the horizontal range of a projectile be a and the maximum height attained by it is b, then prove that the velocity of projection is

If the horizontal range of projectile be (a) and the maximum height attained by it is (b) then prove that the velocity of projection is [ 2 g (b+ a^2 /(16 b)) ] ^(1//2) .

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.

The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now beings to blow in the direction of horizontal motion of projectile, giving to constant horizontal acceleration equal to g. Under the same conditions of projections , the new range will be (g = acceleration due to gravity)

If R is the maximum horizontal range of a particle, then the greatest height attained by it is :

The maximum horizontal range of a projectile is 400 m . The maximum value of height attained by it will be

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

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

An object is projected with speed u and range of the projectile is found to be double of the maximum height attained by it . Range of the projectile is .