V_{i} = Initial velocityV _{f} = Final velocityV _{x} = Horizontal velocity componentV _{y} = Vertical velocity componenta = 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 ( a_{x} = 0). So we can rewrite above equations for horizontal motion as | |||

For the vertical motion, the equations are | |||

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 V_{fy} = 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. |