Asked by Axel
A rock is thrown straight up from a cliff which is 38 meter above ground. After 2 seconds the rock is at 21.3 meter above ground. What was the velocity at first when the rock was thrown. Ignore resistent of the air.
Answers
Answered by
Scott
vi = initial velocity , hi = initial height
h = 1/2 g t^2 + vi t + hi
21.3 = (-4.9 * 2^2) + 2 vi + 38
h = 1/2 g t^2 + vi t + hi
21.3 = (-4.9 * 2^2) + 2 vi + 38
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