Physics midterm problem: Washington coin toss across Rappahanock River

Physics 205A Midterm 1, fall semester 2013
Cuesta College, San Luis Obispo, CA

"Rappahannock River"
flic.kr/p/88bzQv
justin.critzer

A possible basis for the legend for George Washington throwing a dollar coin across the Potomac river is that:

Washington once threw a piece of slate "about the size and shape of a dollar" across the Rappahannock River...[which] measures only [76 m] across, a substantial but perhaps not impossible distance to throw.[*]

Determine whether this coin throw across the Rappahanock River was plausible. Assume a 30° launch angle, and that George Washington could throw a coin with a comparable high school fastball speed[**] of 36 m/s. Neglect air resistance. Show your work and explain your reasoning using properties of projectile motion.

[*] "250 feet," mountvernon.org/content/facts-falsehoods-about-george-washington-0.
[**] "(80 [mph]) 15yo HSV," efastball.com/baseball/pitching/grips/average-pitching-speed-by-age-group/.

Solution and grading rubric:

• p:
  Correct. Concludes that throw was feasible. Finds x- and y-components of initial velocity: v_0x = +31.2 m/s, v_0y = +17.6 m/s; then applies projectile motion equations in one of the following strategies:
  1. Finds t for coin to reach final elevation of y = 0, and determines that the coin would travel to a final horizontal position x = v_0x·t, which is more than 76 m.
  2. Finds required t for coin to travel to a final horizontal position of x = +76 m, and determines that the coin would be above the original elevation by that time.
  3. Finds time to reach highest point in trajectory, then doubles this time and concludes that the coin would travel more than across the river in that time.
  4. Finds that the coin is more than halfway across the river at its highest point.
• r:
  Nearly correct, but includes minor math errors. Correctly solves for the required time for the coin to travel 76 m horizontally, but typically finds the time for the coin to reach its highest point and does not double that time, thus concluding that the coin cannot cross the river.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least enough steps are shown that would theoretically result in a complete answer (such as finding x- and y-components of initial velocity vector, and either time for the coin to travel 76 m horizontally, or reach its highest height, multiple math errors notwithstanding.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at systematic use of kinematic equations for projectile motion (such as finding x- and y-components of initial velocity vector).
• x:
  Implementation of ideas, but credit given for effort rather than merit.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: midterm01p0To
p: 13 students
r: 14 students
t: 14 students
v: 11 students
x: 21 students
y: 0 students
z: 0 students

A sample "p" response (from student 7253), calculating the time for the coin to return to the launch height by multiplying the time to reach its highest height by two, and finding that the horizontal distance traveled would be more than 76 m:

Another sample "p" response (from student 1227), calculating the time for the coin to travel 76 m horizontally, then determining that the coin would be 15 m vertically above its starting height after traveling that horizontal distance: