Show that a projectile covers the maximum horizontal range when angle of projection is `45^@` with horizontal.

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^@ .

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

What is a projectile ? Obtain an expression for: maximum height , time of flight and the horizontal range when a projectile is fired at an angle theta with the horizontal.

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

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

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