# Projectile Motion – Pool ball leaves the table with initial horizontal velocity

A pool ball leaves a 0.20-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.

Vi = Initial velocity
Vf = Final velocity
Vx = Horizontal velocity component
Vy = Vertical velocity component
a = Acceleration
t = time
H = Initial height
X = Horizontal displacement

For the horizontal motion, the equations are

Remember horizontal acceleration is always zero for this problem ( ax = 0). So we can rewrite above equations for horizontal motion as

 Equation (1) Equation (2)

For the vertical motion, the equations are

 Equation (3) Equation (4)

Equation (5)

Apply equation (3) for vertical motion from point A to B

At the initial position A, there is no velocity component on vertical direction, so it is zero Vfy = 0

t = 0.2 s

Pool ball hit the ground in 0.2 seconds.

Apply equation (1) for horizontal motion from point A to B

X = 0.48 m
Horizontal displacement 0.48 meters.

Lets calculate the final velocity and the angle Θ with horizontal pool ball going to hit the ground. Apply equation (4) for vertical motion from point A to B

Calculate final velocity component.

Final velocity on horizontal directions can be easily obtained from equation (2)

Now we can find the final velocity using trigonometry

Final velocity of the pool ball when it hit the ground is 3.098 m/s

Calculate impact angle

Θ = 39.24

Angle of incident 39.24 degrees.