# Ramp angle and landing velocity

1. Jun 10, 2010

### clockworks204

1. A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcycle. There is a 15. m drop and the horizontal distance originally planned was 60. m but it turns out the canyon is really 66.3 m across. If he desires a 3.0-second flight time, what is the correct angle for his landing ramp (deg, positive angle below horizontal) AND landing velocity?

2. magnitude= sqrt(Vf^2 + Vi^2), etc

3. I've calculated the launch speed to be 24.1 m/s and the launch angle to be 23.7 deg. I've tried using the formula 2. to solve for the landing velocity by figuring Vf 15sin(23.7)-(9.81)(3)= -23.4. plugging into the formula, sqrt(-23^2 + 24.1^2)= 33.6 and this is incorrect. I've exhausted myself trying to figure these last 2 parts, and I'm now stumped! Thanks in advance. I also added a visual that may help.

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Last edited: Jun 10, 2010
2. Jun 10, 2010

### rock.freak667

this is correct.

Recheck this as I do not -23.4. Also remember this will give you the vertical component, which must be added (vector wise) to the horizontal component of velocity which is not 24.1 m/s.

Last edited: Jun 10, 2010
3. Jun 10, 2010

### clockworks204

I just rechecked the math and still came out with -23.4. Would the horizontal component be 22.1 from dividing 66.3/3 ? Also, how can the landing angle be calculated? Sorry...confused

4. Jun 10, 2010

### rock.freak667

24.1sin(23.7) - 9.81(3)

yes the horizontal component would be 22.1 m/s.

5. Jun 10, 2010

### clockworks204

Right, wrong number. I figured out the ramp angle by dividing -19.7 by 22.1 and took the reverse tan to get 41.7 degrees. I the got the velocity by using sqrt(19.7^2 + 22.1^2) to get 29.6 m/s.

Thanks for your help rock.freak667, I do appreciate it!