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River Stone
A stone is dropped into a river from a bridge 48.7 m above the water. Another stone is thrown vertically down?
A stone is dropped into a river from a bridge 48.7 m above the water. Another stone is thrown vertically down 1.28 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?
Start with
y =(1/2)gt^2 for the first stone and
solve for the drop time:
t =sqrt(2y/g)=sqrt(2*48.7/9.8) = 3.15 s
Now since the second stone was thrown 1.28 seconds later and struck at the same time, then it took (3.15-1.28)= 1.87 seconds to reach the bottom.
v(average) = distance /time = 48.7 m/ 1.87 s = 26 m/s
But we also know that the average speed is
v(ave) = (vi +vf)/2
26 = (vi +vf)/2
52 = (vi +vf)
vf = 52-vi
Now we also know
vf =vi+gt
vf=vi+(9.8)(1.87)
vf = vi + 18.326
So combining the two
vi + 18.326 =52-vi
2vi = 52 -18.326
vi=16.83700 m/s
I'd appreciate it if you voted this a 5 star answer. When you choose a best answer you get back 3 points.
CLIP FROM RIVER STONE
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