A ball is thrown horizontally from the top of a building 36.7 m high. The ball strikes the ground at a point 91.9 m from the base of the building. The acceleration of gravity is 9.8 m/s 2. Find the x component of its velocity just before it strikes the ground.
We can find the time that it takes the ball to hit the ground as follows :
-36.7 = (1/2) * -(9.8) * t^2
-36.7 = -4.9 * t^2
(36.7 /4.9) = t^2 take the positive root
t ≈ 2.74 sec
The x component of the velocity remains constant an can be found as
x = vx * t
Where x is the horizontal displacement, vx is the horizontal component of the velocity and t is the time it takes the ball to hit the ground....so....
91.9 = vx * 2.74
91,9 / 2.74 = vx ≈ 33.6 m / s