Physics 110 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Chapter 3 Q14
Griffith, W. Thomas; The physics of everyday phenomena: a conceptual introduction for
physics;4th Edition ISBN 0-07-250977-5
THE PROBLEM STATEMENT
Ch3 Q14. A ball is thrown straight upward. At the very top of its flight, the velocity of the ball is
zero. Is its acceleration at this point also zero? Explain
©Hulan E. Jack Jr. Oct.10,2004
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Physics 110 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch3 Q14. A ball is thrown straight upward. At the very top of its flight, the velocity of the ball is
zero. Is its acceleration at this point also zero? Explain
Basic Solution (Minimum Expected from the student)
This is called free fall. The ball is under the influence of gravity during the whole time. Hence it
is always accelerating downward at 9.8 m/s2 on earth. So, the acceleration is NEVER zero during
this trip.
Annotated Solution and Discussions (For better understanding)
The figure shows the velocity vs time graph of the ball, or any other object, that is thrown
straight up with some initial velocity v0 . Up has been chosen as the positive direction. The slope
of the v vs t curve is the acceleration. The slope is
constant for this trip, so the acceleration is constant
with value g = -9.8 m/s2 on earth. ( “-“ 9.8m/s2
because up is + and this is down.) Only the velocity
changes as you approach, or leave the top. As it
approaches the top its upward velocity slows coming
to zero at the top - momentarily stopped. Then it falls
downward faster, faster, faster with increasing
negative (downward) velocity.
©Hulan E. Jack Jr. Oct.10,2004
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Physics 110 Problems - My Solutions
Dr. Hulan E. Jack Jr.
A Detailed Numerical Example (A detailed explanation hopefully to aid your
understanding , not to confuse you. Follow the points A, B, C, etc, carefully)
The picture below shows the details of a ball thrown upward with an initial velocity v0 = 20 m/s
upward.
The left side shows the velocity and acceleration vectors, magnitudes and directions, as the ball
rise “up“ , and as it falls “down” at various heights above the starting height, points A to I.
A (start-bottom), B, C, D up; E at the top; F, G, H, I (bottom again) down.
The graphs on the right show the displacement vs time curve (top graph) and the velocity vs
time curve (bottom graph). They show the numerical values for the points A to I.
The numbers come from v = v0 - gt and y = v0t - ½gt2 , where v0 =20 m/s and g=9.8m/s2 .
©Hulan E. Jack Jr. Oct.10,2004
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