A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcyc

Question

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 67.7 m across. If he desires a 3.0-second flight time, what is the correct angle for his launch ramp (deg)?
I am having a hard time finding out how to get the angle. I think I have determined the x and y components of velocity v(x)=22.57 m/s and v(y)=9.715 m/s but I could be wrong here.

What is his correct launch speed?

What is the correct angle for his landing ramp (give a positive angle below the horizontal)?

What is his predicted landing velocity. (Neglect air resistance.)

Explanation / Answer

range R = 67.7 m time of flight T = 3 s we know R = horizontal component of velocity * T from this horizontal component of velocity = R / T                                              v cos = 22.566             -------( 1) In vertical direction : -------------------- Initial velocity u = v sin distance S = height = 15 m accleration a = -g = -9.8 m / s^ 2 time t = T from the relation S = ut + ( 1/ 2) at^ 2                        15 = v sin *T - ( 1/ 2) gT^ 2                        15 = 3 v sin - 44.1          3v sin = 59.1            v sin = 19.7       -----------( 2) eq( 2) / eq( 1) ==> v sin / v cos = 19.7 / 22.556                                                tan = 0.8729                                                      = 41.11 degrees plug it in eq( 1) we get   v = 22.5666/ cos 41.11                                        = 29.95 m / s (b). time taken to reach the maximum height t = v sin / g                                                                      = 2.01 s So, time taken to reach the ground from maximum height t ' = T - t = 0.9897 s vertical component of velocity when it landing V ' = gt ' = 9.7 m / s horizontal component of velocity when it landing V " = 22.566 m / s Since there is no accleration along horizontal direction So, land ing speed V = [ V'^ 2 + V"^ 2]                                  = 24.56 m / s (c). landing angle = tan -1( V ' / V " )                               = 23.25 degrees

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.