Dirt Bike Jump

## Question:

A dirt bike is seen to jump from a small mound, leaving the
ground at an angle of 60 degrees. It lands 20 feet away on level
ground. What was the approximate speed of the bike just before it
left the ground. Neglect the size of the bike and air resistance.
## Answer:

At a 60 degree take off angle the vertical velocity is
0.866*speed, the horizontal velocity is 0.5*speed. The bike stays
in the air long enough that the horizontal velocity carries it 20
feet, while subject to a downward acceleration of 32f/s/s. It
will go 20 feet in 20/(.5*speed) seconds. In that same time its
vertical velocity will go from .866*speed through zero and to
-.866*speed. The change in vertical velocity or 2*0.866*speed is
going to be 32*t, the time in the air. So from the horizontal
component we get t=20/(0.5*speed). From
the vertical component we get
t=2*0.866*speed/32. Setting t=t we get
20/(0.5*speed)=2*0.866*speed/32. Solve for speed.
I get 20*32=0.866*speed^{2} or
speed^{2}=640/.866=739. So speed = 27.2 feet per second.

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