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The following graph of altitude vs. time shows the positions of a projectile that is in a state of freefall in which the only force acting upon it is the pull of gravity. The projectile was launched directly upwards. Use the information given in the graph to answer the following questions.

General Questions
 How far horizontally from its initial position does the projectile land?
 1

 What is the velocity at point A?
 2

 Is the projectile being accelerated as it passes through point A?
 3

 Our graph of altitude vs. time is a parabola with symmetry about t = 2 seconds. If the altitude at J (1, 14.7) is 14.7 meters, what is its altitude at t = 3 seconds?
 4

 On the graph, the slope of the tangent drawn at point J represents the projectile's instantaneous velocity, vJ. To one decimal place, the lower point on the "tangent triangle" is (0.40, 12.2) and the upper point is (1.2, 20.0). Based on this coordinates, what would be our best estimate of the value of vJ?
 5

 Our graph of altitude vs. time is a parabola with symmetry about t = 2 seconds. If the instantaneous velocity at J (1, 14.7) is vJ, what is its instantaneous velocity at t = 3 seconds?
 6

 How high is position A, the apex?
 7

 What is the total distance traveled by the projectile from its initial release position to it location at t = 3 seconds?
 8

 What is the net displacement of the projectile from its initial release position to it's location at t = 3 seconds?
 9

 We can use the equation to model the motion of our projectile.   For any object in freefall, there is negligible air resistance and we use the value of a = -9.8 m/sec2, or a = -g, for its acceleration.    Using this value for acceleration and the data from our graph and previous answers, find the original velocity (the variable vo in our formula) with which the projectile was initially launched.
 10 AlgebraLAB Project Manager    Catharine H. Colwell Application Programmers    Jeremy R. Blawn    Mark Acton Copyright © 2003-2022 All rights reserved.