The maximum range of projectile is 2/ √ 3 times actual range. What isthe angle of projection for the actual range ?

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

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

Range of a projectile is R, when the angle of projection is 30^(@) . Then, the value of the other angle of projection for the same range is

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

The maximum range of a projectile is 22 m. When it is thrown at an angle of 15^(@) with the horizontal, its range will be-

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 range of a projectile is R when the angle of projection is 40^(@) . For the same velocity of projection and range, the other possible angle of projection is

A body is thrown at an angle alpha to the horizontal with initial velocity v_0 .Assuming air drag to be negligible , find (i)the time of motion, (ii)maximum height of ascent and the horizontal range, (iii)the angle of projection at which they are equal , (iv)the angle of projection at which the horizontal range is maximum and the angle at which the vertical range is maximum.