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Physics quiz question: maximum trajectory height

Physics 5A Quiz 3, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 3.37(a)

[3.0 points.] A ball is thrown with an initial velocity of 24.0 m/s at an angle of 30.0 degrees above the horizontal. What is the ball's highest point in its trajectory, as measured from above its launching point?
(A) 0.612 m.
(B) 7.35 m.
(C) 14.7 m.
(D) 24.0 m.

Correct answer: (B)

At the highest point in its trajectory, the vertical velocity component v_f,y is zero. With the acceleration due to gravity a_y = -9.83 m/s^2, and the initial vertical velocity component v_i,y = v_i*sin(30.0 degrees), the maximum displacement delta_y can be solved for:

v_f,y^2 - v_i,y^2 = 2*a_y*delta_y,

delta_y = -(v_i*sin(30.0 degrees))^2/(2*a_y).

Response (A) is -v_i*sin(30.0 degrees)/(2*a_y); response (C) is -(v_i*sin(30.0 degrees))^2/a_y (the modal response, with the factor of 2 dropped); while response (D) is 2*v_i*sin(30.0 degrees).

Student responses
Sections 4987, 4988
(A) : 0 students
(B) : 19 students
(C) : 21 students
(D) : 3 students

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