Posted by **Mary** on Tuesday, March 20, 2007 at 6:32pm.

A cyclist approaches the bottom of a gradual hill at a speed of 14 m/s. The hill is 4.9 m high, and the cyclist estimates that she is going fast enough to coast up and over it without peddling. Ignoring air resistance and friction, find the speed at which the cyclist crests the hill.

Vf = square root of (V0^2 + 2g (h0 - h1)

Vf = square root of (14 m/s)^2 + 2(9.81)(4.9)

Vf = square root of (196m^2/s^2 + 96.138

Vf = square root of (292.138)

Vf = 17.09 m/s

You've got a sign wrong somewhere. How can you coast uphill and end up faster at the top?

please tell me where i went wrong

(ho - h1) is not a positive number. You have taken it to be +4.9 m

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