Review

[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?

[HallsofIvy]

The horizontal direction??

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