The range of a cannonball fired horizontally from a cliff is equal to the height of the cliff. What is the direction of the velocity vector of the projectile as it strikes the ground? (Ignore any effects due to air resistance.)
Since the distance traveled vertically equals the distance traveled horizontally we can set these two equal to solve for the initial velocity. y = x
y = 1/2(9.80m/s^2)*t^2 = v *t. Divide equation by t (ignoring useless the t=0 solution)
4.9m/s^2 *t = v.
Let's assume that initial velocity is 4.9m/sec horizontal. Then t = 1sec and cliff height = 1/2(9.8)* 1^2 = 4.9 meters. We can pick an arbitrary velocity because if problem has solution, any combination velocity and time work as long as 4.9m/s^2 * t = v.
Calculate the final vertical from mechanical energy conservation. KE + ME = KE(final)
1/2(m)4.9^2 + m(9.8)*4.9 = 1/2mv^2. masses can be canceled
60 = 1/2v^2. V = 10.95. But horizontal component is 4.9. So direction is cos -1 (4.9/10.95) or 63.4 degrees below horizontal.
As a check:
Try with initial velocity = 9.8m/sec. Time to fall = 2s, and cliff height = 1/2(9.8)*4 = 19.6m
Using energy balance again
1/2(m)(9.8)^2 + m(9.8)(19.6m) = 1/2 m V^2.
48+192 = 1/2 V^2. V = 21.91m/s. Again angle = cos -1(9.8/21.91) = 63.4 degrees below horizontal.