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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 | |||
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Remember horizontal acceleration is always zero for this problem ( ax = 0). So we can rewrite above equations for horizontal motion as | |||
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For the vertical motion, the equations are | |||
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![]() Equation (5) | |||
Apply equation (3) for vertical motion from point A to B | |||
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At the initial position A, there is no velocity component on vertical direction, so it is zero Vfy = 0 | |||
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t = 0.2 s | |||
Pool ball hit the ground in 0.2 seconds. | |||
Apply equation (1) for horizontal motion from point A to B | |||
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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 | |||
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Calculate final velocity component. | |||
Final velocity on horizontal directions can be easily obtained from equation (2) | |||
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Now we can find the final velocity using trigonometry | |||
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Final velocity of the pool ball when it hit the ground is 3.098 m/s | |||
Calculate impact angle | |||
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Θ = 39.24 | |||
Angle of incident 39.24 degrees. |