Answer to Question #139305 in Classical Mechanics for Gab

Question #139305
In a tv show Darna, a stuntwoman drives horizontally with a speed of 1.34 m/s off a platform that is 12.0 m above the water. What is her speed just before striking the water?
1
Expert's answer
2020-10-22T18:07:23-0400

In order to safe the conservation law:

v=(v2+2gh)12=((1,34)2+2×9,8×1,2)12=5.03v' = (v^2 + 2gh)^\frac{1}{2} = ((1,34)^2 + 2\times9,8\times1,2)^{\frac{1}{2}} = 5.03


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