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 `[2g(b+a^2/(16b))]^(1/2)`

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 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 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)

The horizontal range of a projectile is R and the maximum height at tained by it is H. A strong windnow begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now be:

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

A ball is projected to attain the maximum range. If the height attained is H, the range is