A motorcyclics performing a stunt - How long is he in the air?

1. Mar 15, 2004

celect

A motorcyclist intends to perform a stunt in which he jumps a 25meter wide river.

The motorcyclist must jump from a cliff on one-side 20 meters lower than the other side.
The motorcyclist approaches the cliff at 50m/s

What formula should I use to solve for

the Horizontal direction?

And the time in the air?

2. Mar 15, 2004

HallsofIvy

Staff Emeritus
The horizontal direction??

What you want to do is set up the equations for distance covered with unknown angle &theta; and initial speed 50 m/s. The vertical acceleration is -9.8 m/s2 so the vertical speed at any t is 50 sin(&theta;)- 9.8t and the vertical height above the starting point is 50 sin(&theta;)t- 4.9t2. Since the landing side is 20 m higher than the starting side, the jump should end with
50 sin(&theta;)t- 4.9t2= 20. Also his horizontal distance covered is 50 cos(&theta;)t. In order to get across the river, he needs to have 50 cos(&theta;)t= 25. Solve the two equations for t and &theta; to answer your question.