Question 30576
There are two eqns of motion. One for the ball falling down. And one for the ball which is thrown upwards.
Let s1 represent the height of the dropped ball, above the ground
Let s2 represent the height of the thrown ball, above the ground.
Using the eqn of motion you have, S = -16t2 + v0t + s0, and substituting in appropriate values for s0 and v0, (upwards is taken as positive)
s1 = 60 - (1/2)gtē
s2 = 80t - (1/2)gtē
subtracting one height from t'other,
s1 - s2 = 60 - 80t
For the difference in height to be less than or equal to 10, we can write,
|s1 - s2| = |60 - 80t| <= 10
We therefore have two inequalities,
60 - 80t <= 10
and
80t - 60 <= 10
From the 1st inequality,
60 - 80t <= 10
60 - 10 <= 80t
80t >= 50
t >= 5/8
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From the 2nd inequality,
80t - 60 <= 10
80t <= 70
t <= 7/8
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The balls will be less than or equal to 10 feet apart in the time interval denoted by.
5/8 <= t <= 7/8
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