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?.
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?)
?y= (43.3m/s)(sin 12?)(1.8s)+.5(-9.81m/s/s)(1.8s)^2
Not sure if this is right...