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you throws two stones from the top edge of a building with a speed of 48.0 m/s. you throw one straight down and the other straight up. the first one hits the street in a time t1. how much later is it before the second stone hits?



Answer :

Answer:

H1 = V t1 - 1/2 g t1^2      where upward is chosen as positive

H2 = -V t2 - 1/2 g t2^1     second stone thrown downwards

V t1 - 1/2 g t1^2 = -V t2 - 1/2 g t2^2

V (t1 + t2) = 1/2 g (t1^2 - t2^2)

V = 1/2 g (t1 - t2)

t2 = 2 V / g = 96 / 9.8 = 9.8 sec

Check:

One can also note the stone 1 will return to the starting point with a time delay of V / g * 2 which is 96 / 9.8 = 9.8 sec because the time for stone 1 to go to zero is 48 / 9.80 = 4.90 sec when its velocity reaches zero

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