How High Does the Daredevil Reach?

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Homework Help Overview

The problem involves a daredevil on a motorcycle launching off a ramp with an initial speed and reaching a peak height with a different speed. The context is related to kinematics and projectile motion, focusing on the maximum height achieved during the jump.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the speeds at the ramp and peak, questioning how to derive the angle of projection and its implications for calculating the vertical component of velocity.

Discussion Status

Some participants have suggested methods to find the angle of projection and are exploring how to use that angle to determine the vertical component of velocity. There is ongoing dialogue about the next steps in the calculation process, with no clear consensus yet.

Contextual Notes

Participants are working under the assumption that air resistance and friction can be ignored, and there is a focus on the relationship between horizontal and vertical components of motion.

Aznhmonglor
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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.
 
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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?
 
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:
Aznhmonglor said:
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?

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.)
 

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