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? (b) At what height do they hit each other? (c) redo part (a) and part (b), with the assumption that the tennis ball is thrown 1.00 s before the baseball.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 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?
=====================================
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
-------
-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.
========================
(b) At what height do they hit each other?
n(1.39888) =~ 7.198 meters
----------------------------------
(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.
|
|
|
