Maximum height of a projectile is H an range of projection is R when the projectile is projected with the velocity u. Show that `R^2 = 16H (u^2/(2g)-H)`.

The maximum height attained and the range of a projectile are H and R respectively. If projected with an initial velocity of u, show that, R^2=16H((u^2)/(2g)-H).

What are two angles of projection of a projectile projected with velocity 30 m/s , so that the horizontal range is 45m . Take g=10 m/s^2 .

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

Find the angle of projection of a projectile for which the horizontal range and the maximum height are equal.

The equation of the trajectory of a projectile on a vertical plane is y= ax-bx^2 , where a and b are constants, and x and y respectively are the horizontal distances of the projectile from the point of projection. Find out the maximum height attained by the projectile, and the angle of projection with respect to the horizontal.

The velocity of a projectile is 15m/s .At what angle to the horizontal should it be projected so that it cover maximum horizontal distance?

A projectile is launched from the ground and it returns to the ground level. The horizontal range of the projectile is R= 175 m . If the horizontal component of the projectile's velocity at any instant is 25 m *s^(-1) , then determine the time of flight of the projectile.

What is a projectile .Show that the path following by a projectile is a parabolic , when it is projected at an angle theta with the horizontal. Obtain an expression for maximum height.

What is a projectile .Show that the path following by a projectile is a parabolic when it is projected at an angle theta with the horizontal. Obtain an expression for Horizontal range.