A Hollywood Daredevil plans to jump a canyon 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 69.6 m across. If he desires a a 3.2 second flight time
a. what is the correct angle of his launch ramp?
b.what is his correct launch speed?
c.what is the correct angle for the landing ramp?
d.what is his predicted landing velocity?
4 answers
I answered this question last night. Have you changed screen names? The 60 m "original" information is irrelevant
yes, same person I just don't understand why im getting this problem wrong for a. predict the launch angle. I do tan inverse of 7.87/21.75 giving an angle of 19.9 degrees so I tried subtracting from 180 degrees but im not getting it right either way can you explain please?
consider the horizontal
69.6=speed*cosTheta*3.2
consider the vertical
-15=speed*sinTheta*3.2- 4.9 3.2^2
solve each side for speed first.
speed= 69.6/(3.2cosTheta)
speed= (-15+4.9*3.2^2)/sintheta*3.2
set them equal
69.6/(3.2cosTheta)=(-15+4.9*3.2^2)/sintheta*3.2
multipy both sides by sinTeta
69.6/3.2 * tanTheta= (-15/3.2 + 4.9*3.2)
solve for Theta.
tan theta= (-4.68+15.68)/21.75=.505
I get 26 deg. Check my work and thinking carefully. Very carefully.
69.6=speed*cosTheta*3.2
consider the vertical
-15=speed*sinTheta*3.2- 4.9 3.2^2
solve each side for speed first.
speed= 69.6/(3.2cosTheta)
speed= (-15+4.9*3.2^2)/sintheta*3.2
set them equal
69.6/(3.2cosTheta)=(-15+4.9*3.2^2)/sintheta*3.2
multipy both sides by sinTeta
69.6/3.2 * tanTheta= (-15/3.2 + 4.9*3.2)
solve for Theta.
tan theta= (-4.68+15.68)/21.75=.505
I get 26 deg. Check my work and thinking carefully. Very carefully.
ok thanks very much I understand the upward motion part of the problem. I am still have trouble with the downward motion can someone explain. Thanks for the help
1 answer