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ramp angle and landing velocity

 
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Jun10-10, 07:14 PM   #1
 

ramp angle and landing velocity

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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Jun10-10, 07:42 PM   #2
 
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Quote by clockworks204 View Post
3. I've calculated the launch speed to be 24.1 m/s and the launch angle to be 23.7 deg.
this is correct.

Quote by clockworks204 View Post
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.
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.
Jun10-10, 07:55 PM   #3
 
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
Jun10-10, 08:09 PM   #4
 
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ramp angle and landing velocity

Quote by clockworks204 View Post
I just rechecked the math and still came out with -23.4.
24.1sin(23.7) - 9.81(3)

Quote by clockworks204 View Post
Would the horizontal component be 22.1 from dividing 66.3/3 ? Also, how can the landing angle be calculated? Sorry...confused
yes the horizontal component would be 22.1 m/s.
Jun10-10, 08:39 PM   #5
 
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!
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