A driver runs horizontally with a speed of 1.20 m/s off a platform that is 10.0 m above the water.?
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what is his speed just before striking the water?
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Answer:
I would use the formula v^2 - u^2 = 2as. Where v is the final speed, what you are looking for. u is the initial speed, 1.2, a is gravity, 9.81m/s^2, s is 10m. Therefore: v^2 - u^2 = 2as v^2 - 1.2^2 = 2 x 9.81 x 10 v^2 - 1.44 = 196.2 v^2 = 196.2 + 1.44 v^2 = 197.64 v = sqrt 197.64 v = 14.06 m/s Hope this helps.
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