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## Homework Statement

At the base of a frictionless icy hill that rises at 28.0 degrees above the horizontal, a toboggan has a speed of 12.0m/s toward the hill. How high vertically above the base will it go before stopping?

## Homework Equations

height = [(v*sin (angle))^2] / (2g)

## The Attempt at a Solution

height = [(v*sin (angle))^2] / (2g)

height = [(12m/s*sin(28degrees))^2] / (2(9.8m/s^2))

height = 31.74m^2/s^2 / 19.6m/s^2

height = approx 1.62m

I think I made a mistake with the velocity because this answer is incorrect. Can someone help me please and thank you.