Question

Click on the variable x, and change it to y.  Paste a snapshot below and explain how it is different than the plot of x and why that is so. Here is the snapshot and it is a parabola. The plot of x is here:

76% [5), QE i Tracker Wed 1101 PM Tracker File Edit Video Track Coordinate System View Help v projectile mo memory in use: 26After changing x position to y: picture is below. It is a parabola. This has to do with constant acceleration?

令42% Thu 11:55 AM Q E i Tracker Tracker File Edit Video Track Coordinate System View Help ▼ O projectile mr memory in use: 27

76% [5), QE i Tracker Wed 1101 PM Tracker File Edit Video Track Coordinate System View Help v projectile mo memory in use: 26MB of 228MB 90Track Con.. Plot projectile model O projectile model projectile model (t, x) 1.0 0.8 0.6 0.4 0.2 0 0.1 0.2 0.3 0.4 0.50.6 2 (s) 1-0.693 s x-1.195 EE Tableprojectile model y (m) 0.000 8.127E-2 x (m) t(s) 0.000 0.000 0.033 5.689E-2 0.066 0.114 0.152 0.099 0.212 0.228 0.261 0.132 0.165 0.284 0.300 0.198 0.341 0.328 St 0.231 0.398 0.345 0.351 Mo 0.347 0.332 Pan 0.455 0.264 0.297 0.512 X1.387 m y-8.882E-2 m 0.330 0.569 0.626 030 100%に 0.363 0.307 0.396 0.683 0.271 projectile model.trk
令42% Thu 11:55 AM Q E i Tracker Tracker File Edit Video Track Coordinate System View Help ▼ O projectile mr memory in use: 27MB of 228MB Track Con Plot projectile model O projectile model projectile model (t, y) 0.2 0 2 -Auto- -5.34E-1 0 0.1 0.2 0.3 0.4 0.5 0.6 t-0.693s y -0.534 m projectile model Table t (s) x (m)y (m) 0.693 195 -0.534 1.138 -0.40 0.627 1.081-0.28 1.024 0.170 0.594 0.561 0.967 -6.982 0.000 0.0000.000 0.9101.966E-2 5.689E-28.127E-2 0.528 0.853 9.847E-2 0.114 0.152 0.462 0.796 0.167 0.099 0.171 0.212 030 100% 0.7400.224 projectile model.trk
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Answer #1

The object has no acceleration in x direction, hence the velocity of the object in x direction does not change and the x vs t graph has linear shape.

But the object has acceleration in y(downward) direction and its constant, hence y vs t graph is parabolic in shape.

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