Question
You are the physics expert for a professional stuntman. The stunt you have designed includes a ramp that is angled at 30°. You expect that the stuntman will be traveling at 18.0 m/s when he leaves the ramp. How far must the landing ramp be positioned, and what is the maximum height of the flames he will be jumping over?
Answers
Asuuming the landing ramp is at the same elevation at the takeoff ramp,
Range = (V^2)/g * sin 60 = 38.2 m
maximum height = (V sin30)^2/(2 g) = V^2/8g = 4.1 m.
Range = (V^2)/g * sin 60 = 38.2 m
maximum height = (V sin30)^2/(2 g) = V^2/8g = 4.1 m.
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