HELP PLEASE!!! A pool ball leaves a 60 cm 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.

Question

anonymous · Answer

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For elapsed time to ball hit the ground, you can use $$Y= V _{y}*t + \frac{ g*t ^{2} }{ 2 }$$

Where, Vy equals zero. 

Then, for Range R  

$$R=V _{x}*t $$

where t is elapsed time that you have founded before.

anonymous · Answer

What does the Y, g and t^2 stand for? :(

anonymous · Answer

you'll want to use the kinematic equation of:
y = y +vt+(1/2)gt^2
the initial height of the ball is 60cm or 0.6m
y = 0.6 +vt + (1/2)gt^2
the final height of the ball will be zero, for ground level
0 = 0.6 +vt + (1/2)gt^2
the initial velocity in the y-direction of the ball is zero because the only reason it will fall is from gravity.
0 = 0.6 +(0)t + (1/2)gt^2
g stands for gravity, which is what accelerates the ball down. it is always equal to -9.8 when you have units of 'meters'
0 = 0.6 +(0)t + (1/2)(-9.8)t^2
now all that is left is for you to solve for t, which is time.
does this all make sense where all the numbers go? @peekaboopork

anonymous · Answer

Yes, thank you so much!