Relation between limit or Range of projection (R) & flight time is` R = 5T^2`. Calculate the angle of projection.

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

Statement I: The horizontal range of a projectile is the same for angles 30 ^@ and 60^@ of projection. Statement II: The horizontal range of projectile is independent of the angle of projection.

If the horizontal range is 4 times of the maximum height then what is the angle of projection?

The equations of motion of a projectile are x=36 t and 2y=96t- 9.8 t^2 . The angle of projection is

Statement I: The initial velocity of projectile =(ahati+bhatj) . The horizontal range becomes maximum for a=b. Statement II: for the same magnitude of initial velocity , the horizontal range of a projectile becomes maximum for the angle 45^@ of projection.

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.

For a projectile, ("horizontal range")^2 =48xx("maximum height")^2 The angle of projection is

If a projectile has a constant initial speed and angle of projection, find the relation between the changes in the horizontal range due to change in acceleration due to gravity.

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 ?

If the relation between two variable x and y is 5y+7x=11 and the range of x is 5, then the range of y is _______ .