20181012

Physics midterm problem: plausible cliff height for ATV jump

Physics 205A Midterm 1, fall semester 2018
Cuesta College, San Luis Obispo, CA

In 2005, Chip Gaines reportedly drove a four-wheeled all-terrain vehicle over an embankment at the edge of a cliff, and became airborne:
I gunned it and launched that four-wheeler straight off the other side of the hill—over a sheer cliff that dropped a good twenty feet to the ground... In a matter of two seconds, the four-wheeler and I...face-planted into the dirt from nearly twenty feet up... And that's how I wound up with this awesome scar.[*]
While Chip Gaines claims that the cliff was 20 ft high (6.0 m), his wife Joanna Gaines recalls that the cliff was only 10 ft high (3.0 m).

Determine which cliff height (6.0 m or 3.0 m) was more plausible for Chip Gaines to be airborne for two seconds after launching himself on his four-wheeler (presumably a 2003 Kawasaki KVF 360 4⨉4[**]) with a speed of 38 mph (17 m/s) at an angle of 30° above the horizontal. Neglect air resistance, and treat Chip Gaines as a point object. Show your work and explain your reasoning using properties of projectile motion.

[*] Chip Gaines, Capital Gaines: Smart Things I Learned Doing Stupid Stuff, W Publishing (2017), pp. 44-47.
[**] kawasakimotorcycle.org/forum/kawasaki-atv-mule/30810-top-speed-360-a.html.

Solution and grading rubric:
  • p:
    Correct. From the initial speed of v0 = 17 m/s and direction of 30° above the horizontal, finds the y-component of initial velocity v0y = v0·sinθ = +8.5 m/s; applies projectile motion equations to determine that at t = 2 s, Chris Gaines would be at a final height of y = –2.6 m (thus 2.6 m below his starting point of y0 = 0), thus making Joanna Gaines' estimate for the cliff height (3.0 m) more plausible that Chris Gaines' estimate of 6.0 m. May instead started with y = –3.0 m and y = –6.0 m and used the quadratic equation to solve for the expected times to reach the bottom of these cliffs, and found that the time to reach a final position of y = –3.0 m is closer to the given t = 2 s.
  • r:
    Nearly correct, but includes minor math errors.
  • t:
    Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least some systematic attempt at using kinematic equations for projectile motion.
  • v:
    Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at systematic use of kinematic equations for projectile motion.
  • x:
    Implementation of ideas, but credit given for effort rather than merit. No clear attempt at kinematic equations for projectile motion. Primarily applies trigonometry to find distances rather than velocity components.
  • y:
    Irrelevant discussion/effectively blank.
  • z:
    Blank.
Grading distribution:
Sections 70854, 70855
Exam code: midterm01g4iN
p: 31 students
r: 8 students
t: 8 students
v: 3 students
x: 7 students
y: 0 students
z: 1 student

A sample "p" response (from student 0517):

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