volleyball11
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- Oct 30, 2005
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A ball is dropped from 100 feet. Exactly one second later a ball is dropped from 75 feet. Which ball will hit the ground first?
Is this easy?
- Thread starter volleyball11
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All objects accelerate towards Earth at approximately 9.8 m.s<sup>-1</sup> (ignoring air resistance).
Distance travelled in one dimension is given by:
s = v<sub>0</sub>t + (1/2)at<sup>2</sup>
Where v<sub>0</sub> = initial velocity (zero in this case)
t = time
a = acceleration
Therefore, by rearranging the equation,
\(\displaystyle t = \sqrt(2s/a)\)
s<sub>1</sub> = 100 feet = 30.48 m
\(\displaystyle t_1 = \sqrt(2 \times 30.48 / 9.8 ) = 2.494 s\)
s<sub>2</sub> = 75 feet = 22.86 m
\(\displaystyle t_2 = \sqrt(2 \times 22.86 / 9.8 ) = 2.160 s\)
But since the second ball is dropped a second later, if t = 0 is when the first ball is dropped, the second will hit the ground after t<sub>2</sub> + 1 s
Therefore:
First ball hits ground after 2.494 s
Second ball hits ground ater 3.160 s
Obviously the first ball hits the ground first.
Distance travelled in one dimension is given by:
s = v<sub>0</sub>t + (1/2)at<sup>2</sup>
Where v<sub>0</sub> = initial velocity (zero in this case)
t = time
a = acceleration
Therefore, by rearranging the equation,
\(\displaystyle t = \sqrt(2s/a)\)
s<sub>1</sub> = 100 feet = 30.48 m
\(\displaystyle t_1 = \sqrt(2 \times 30.48 / 9.8 ) = 2.494 s\)
s<sub>2</sub> = 75 feet = 22.86 m
\(\displaystyle t_2 = \sqrt(2 \times 22.86 / 9.8 ) = 2.160 s\)
But since the second ball is dropped a second later, if t = 0 is when the first ball is dropped, the second will hit the ground after t<sub>2</sub> + 1 s
Therefore:
First ball hits ground after 2.494 s
Second ball hits ground ater 3.160 s
Obviously the first ball hits the ground first.