Janet jumps off a high diving platform with a horizontalvelocity of 2.49 m/s and

Question

Janet jumps off a high diving platform with a horizontalvelocity of 2.49 m/s and lands in the water at 1.9 s later. The acceleration of gravity is 9.8 m/s2. a) How high is the platform? Answer in units of m. b) How far from the bas of the platform does she land? Answer in units of m. Janet jumps off a high diving platform with a horizontalvelocity of 2.49 m/s and lands in the water at 1.9 s later. The acceleration of gravity is 9.8 m/s2. a) How high is the platform? Answer in units of m. b) How far from the bas of the platform does she land? Answer in units of m.

Explanation / Answer

Y = Yo +VoyT + 1/2at^2 where a = -9.8 m/s^2 Voy = 0 , because he jumps horizontally , no initial yvelocity component... y = 0 + 1/2(-9.8)(1.9s)^2 y = 17.7 m high platform D = Vx*T Vx in constant, no horizontal acceleration D = 2.49m/s*(1.9 s) D = 4.73 m away from platform

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