Projectile motion motorcycle jump

  • #1
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Homework Statement


The current world-record motorcycle jump is 77.0 m,
set by Jason Renie. Assume that he left the take-off ramp at
12.0º to the horizontal and that the take-off and landing
heights are the same. Neglecting air drag, determine his take-off
speed

Homework Equations


R = [(v0)^2/g]*sin2(theta)
5 constant acceleration equations

The Attempt at a Solution


Hi. I understand that I can use the range equation to solve this question, but I wanted to try and use the constant acceleration equations to try and understand it a little better, however I'm getting stuck.

Let u = the take off speed (magnitude of velocity vector)

u_x = ucos(12)
x_i = 0
x_f = 77
a_x = 0
t = ?

u_y = usin(12)
a = -9.8
y_i = 0
y_f = 0

I get stuck here because I don't know the time or final velocity components (not really sure how to use v_f = 0 for half the flight time either)
 

Answers and Replies

  • #2
PeroK
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Let u = the take off speed (magnitude of velocity vector)

u_x = ucos(12)
x_i = 0
x_f = 77
a_x = 0
t = ?

u_y = usin(12)
a = -9.8
y_i = 0
y_f = 0

I get stuck here because I don't know the time or final velocity components (not really sure how to use v_f = 0 for half the flight time either)
You need to think a move or two ahead. These problems don't always come out in one step. You put ##t = ?## but you should keep going with ##t = \frac{x_f}{u \cos \theta}##

Now, try to find a second expression for ##t## using the vertical component. You can do it either by using the time to the maximum height (where ##v_y = 0##) and symmetry of the motion. Or, you could use ##s_y = u_yt + \frac{a_y t^2}{2}## for vertical motion.
 
  • #3
16
0
With your help I was able to solve it! I'll try to think further ahead for future problems.

Thanks so much!
 
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