Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 3.42(c)
[20 points.] A beanbag is thrown with a velocity of 15 m/s at 45° below the horizontal from a window a height 30 m above the ground. At what horizontal distance from the point directly below the window will the beanbag hit the level ground? Neglect air resistance. Show your work and explain your reasoning.
Solution and grading rubric:
- p = 20/20: Correct.
Breaks the initial velocity vector in separate x- and y-components. Note that ax = 0, and ay = -9.80 m/s2. May either use quadratic equation from ∆y = viy·∆t + (1/2)·ay·(∆t)2, to solve for time; or vfy2 - viy2 = 2·ay·∆y and then plugging in vfy into vfy = viy + ay·∆t to find time. Once time is known, it can then be used to find the horizontal displacement ∆x when the beanbag hits the ground. - r = 16/20:
Nearly correct, but includes minor math errors. Otherwise carries out systematic decomposition into x and y-components, to kinematic equations, to reduction/substitution using algebra approach. - t = 12/20:
Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. - v = 8/20:
Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Still has a methodical approach based on the kinematic equations of motion. - x = 4/20:
Implementation of ideas, but credit given for effort rather than merit. May estimate distance from by appealing to trigonometry and straight-line travel. - y = 2/20:
Irrelevant discussion/effectively blank. - z = 0/20:
Blank.
p: 7 students
r: 6 students
t: 5 students
v: 19 students
x: 6 students
y: 0 students
z: 0 students
A sample of a "p" response (from student 2325) using the quadratic formula to first determine the time for the beanbag to hit the ground:
A sample of a "p" response (from student 3153) instead solving for the final vertical velocity of the beanbag in order to solve for the time for the beanbag to hit the ground:
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