The resultant velocity of a projectile at its highest point is `sqrt(6/7)` times that at half the maximum height. Show that the angle of projection with respect to the horizontal is `30^@`.

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

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.

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 ?

Show that the maximum height attained by a projectile is one fourth of its maximum horizontal range.

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

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

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?

In case of a projectite , speed of the particle at the top most position is half of its initial speed . The angle of projection with horizontal plane is

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