Question 659117
A baseball is thrown upward with an initial speed of 12.0 m/s. Exactly 1.00 s later, a tennis ball is thrown vertically along the the exact same path with a speed of 20 m/s (a) at what time they hit each other?
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For the baseball:
b(t) = -4.9t^2 + 12t
Tennis ball:
n(t) = -4.9(t-1)^2 + 20(t-1) = -4.9t^2 + 29.8t - 24.9
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-4.9t^2 + 12t = -4.9t^2 + 29.8t - 24.9
17.8t = 24.9
t =~ 1.39888 seconds after the launch of the tennis ball.
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(b) At what height do they hit each other?
n(1.39888) =~ 7.198 meters
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(c) redo part (a) and part (b), with the assumption that the tennis ball is thrown 1.00 s before the baseball.
There's no "intersection."  No collision.