Asked by Majiri
A ball is thrown upward with speed 12m/s from the top of a building. How much later must a second ball be dropped. From the same starting point if it is to hit the ground at the same time as the first ball? The initial position is 24m above the ground
Answers
Answered by
Steve
1st ball:
24+12t-4.9t^2 = 0
t = 3.75
2nd ball:
24-4.9t^2 = 0
t = 2.21
so, let 1.54 seconds pass
24+12t-4.9t^2 = 0
t = 3.75
2nd ball:
24-4.9t^2 = 0
t = 2.21
so, let 1.54 seconds pass
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