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## Homework Statement

In 1991, Doug Danger rode a motorcycle to jump a horizontal distance of 76.5m. Find the maximm height of the jump if his angle with respect to the ground was 12.0?.

## Homework Equations

Initial Velocity= ?g?x/2(sin ?)(cos ?)

?t= ?x/vi(cos ?)

?y= vi(sin ?)?t+1/2g(?t)^2

## The Attempt at a Solution

vi= ?(9.81m/s/s)(76.5m)/2(sin 12?)(cos 12?)

vi= 43.3 m/s

?t= 76.5m/(43.3m/s)(cos 12?)

?t= 1.8s

?y= (43.3m/s)(sin 12?)(1.8s)+.5(-9.81m/s/s)(1.8s)^2

Not sure if this is right...

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