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P109. Motion of a Horizontally-Launched Projectile

The purpose of the following exercise is to examine the motion of a projectile using a graphical method and to practice vector operations. Present your solutions as described in the instructions below.

The following information is given about the projectile. Positive velocities are to the right and up.

Horizontal Vertical
xo = 0 yo = 0
vox= 0.750 m/s voy = 0.000 m/s
ax = 0.00 m/s² ay = -9.80 m/s²   

A. Equations of motion

  1. Write the complete dvat equations for vx, vy, x, and y. See Table 4-1 in the text.

  2. Substitute into the equations quantities that are zero and rewrite the 4 equations in simplified form.

B. Data: Use the 4 simplified equations to fill in values in a table like the following.

t
(s)

x
(m)

y
(m)

vx
(m/s)

vy
(m/s)

0.000

0.000

0.000

0.750

0.000

0.050

       

0.100

       

0.150

       

0.200

       

C. Graphical construction: For the following, accurate measurement and construction is important in order to obtain good results.

  1. The graph paper provided for this class is ruled in major divisions of centimeters. Plot the 5 positions of the projectile above at actual scale. That is, one centimeter on the graph paper will represent one real-world centimeter of distance. Turn the paper in portrait orientation in order to fit the five points on the grid.

  2. At each position, draw vectors to represent the horizontal and vertical components of the velocity. Draw all vectors to a scale such that 1.0 cm on the graph paper represents 0.50 m/s. Note that this is a distance-to-speed scale, since you’re using distances to represent the magnitude of the velocity.

  3. Use the graphical method of vector addition to construct the velocity vector, v, for each pair of velocity components. See this animation in order to see how your vectors may appear. (You can change the launch angle to 0, but note that the velocity magnitudes are different than given for this problem.)

D. Acceleration vector: The average acceleration vector can be found as the vector difference of two velocity vectors divided by the time interval between them. That is,

.

  1. Select two such velocity vectors and construct their difference in the unused lower left-hand corner of your graph paper. Select the initial velocity vector as one of the two. Label each vector in your vector difference drawing. See Fig. 3-14 on page 64 of your text to review how to construct the difference of two vectors.

  2. Once you've constructed the vector difference, measure its length and use the scale factor to convert to m/s. Then divide by the time interval to find the magnitude of the average acceleration. State both the magnitude and direction of the average acceleration vector.

E. Question: Does the value of average acceleration that you calculated depend on what two velocity vectors you choose? Explain. 



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