Show that the maximum height reached by a projectile launched at an angle of `45^@` is one quarter of its range.

If R is the horizontal range for theta inclination and h is the maximum height reached by the projectile, show that the maximum ragne is given by (R^(2))/(8h) + 2h .

If the maximum vertical height and horizontal ranges of a projectile are same, the angle of projection will be

If the maximum vertical height and horizontal ranges of a projectile are same, the angle of projection will be