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

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 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) .

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) .

A projectile is projected from ground such that the maximum height attained by, it is equal to half the horizontal range.The angle of projection with horizontal would be

A projectile is projected from ground such that the maximum height attained by, it is equal to half the horizontal range.The angle of projection with horizontal would be

A javelin is thrown with an initial velocity v_0 at an angle theta with the horizontal. Find the angle of projection at which the maximum height attained by the javelin is equal to the horizontal range.

In the question number 62. the maximum height attained by the ball is

If the range of a projectile, projected from the surface of the earth is four times the maximum height attained, what is the angle of projection ?