A ball is thrown from the top of a hill at a velocity v and at an angle A to the horizontal. Is there a formula to find its range.
It is not possible to determine the range of the ball if it is thrown from the top of a hill if the height of the hill is not given.
A formula can be derived for the range of a ball on a horizontal surface when it is thrown from ground level at a velocity v and at an angle A with the horizontal.
First, find the horizontal and vertical components of the velocity vector. The horizontal component is v*cos A and the vertical component is v*sin A. There is no acceleration acting on the ball that changes the horizontal component of the velocity. The acceleration of 9.8 m/s^2 acting vertically downwards affects only the vertical component of velocity. As the ball rises the vertical component of the velocity decreases and it is 0 at the highest point. The time taken for the ball to reach the highest point is equal to the time taken for it to come down to ground level again. Use the relation v = u + a*t, that gives the final velocity v of an object after t seconds that starts with an initial velocity u an there is a constant acceleration a acting on it.
As the vertical component of the velocity of the ball is 0 at the highest point, 0 = v*sin A - g*t which gives t = v*sin A/g, where g is the acceleration due to gravity.
The ball is in motion for twice the time t.
In this duration, the horizontal distance travelled by it is 2*t*v*cos A or substituting v*sin A/g for t gives the range of the ball as 2*v*sin A*v*cos A/g
The formula for the range of a ball thrown at an angle A with the horizontal at v m/s is D = v^2*sin 2A/g.