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An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 19.0 m if her initial speed is 3.80 m/s. What is the freefall acceleration on the planet?
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mike

Solution:
To have maximum range for a given initial velocity, her launch angle must be q0 = 45o. Her range then is R = (v02sin2q 0)/g' = v02sin90o/g' = v02/g'. We have g' = v02/R = 0.96m/s2.
posted by Anonymous

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