Friday

December 19, 2014

December 19, 2014

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

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

**Answer this Question**

**Related Questions**

Physics - A cyclist approaches the bottom of a gradual hill at a speed of 14 m/s...

Physics - A cyclist approaches the bottom of a gradual hill at a speed of 20 m/s...

physics - A cyclist approaches the bottom of a gradual hill at a speed of 12.4 m...

Physics - A cyclist approaches the bottom of a gradual hill at a speed of 20 m/s...

Physics - A cyclist starts from rest and coasts down a 6.5∘{\rm ^\circ} ...

physics - A cyclist has a final velocity of 30 m/s after coasting down hill for...

Physics - A cyclist coasts up a 10.3° slope, traveling 17.0 m along the road to ...

Physics - A bicyclist is coasting straight down a hill at a constant speed. The ...

Physics - A bicyclist is coasting straight down a hill at a constant speed. The ...

physics - How fast must a cyclist climb a 4.0 degree hill to maintain a power ...