# Daredevil Height Problem

1. Mar 2, 2009

### Aznhmonglor

A daredevil on a motorcycle leaves the end of a ramp with a speed of 41.0 m/s as in the figure below. If his speed is 39.1 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance.

Help would be appreciated, not sure how to start this problem.

2. Mar 3, 2009

### rl.bhat

At the peak of the path, the velocity is horizontal. It is given. Equate this with horizontal component of the velocity of projection. That gives you the angle of projection.
Can you proceed further?

3. Mar 3, 2009

### Aznhmonglor

Does that mean I get this.

39.1=41cos(theta)

Solve for theta and I get the angle to be 17.5.

What do I do next? Do I do the vertical component of velocity of projection with the new angle?

Last edited: Mar 3, 2009
4. Mar 3, 2009

### LowlyPion

What is the vertical component of velocity then?

Figure the problem now like you had someone throw a ball at that speed straight up. What would the max height be in that case? (The answer is the same for your problem.)