For example, Problem 37 would show downward arrows of increasing length as the melon descends. The space between successively lower points would become larger. In problem 40 the problem with the wad sticking to the ceiling required you to show a series of vertical arrows depicting the motion of the wad as it rose toward the ceiling. Since the arrows indicate the velocity, each arrow should be vertical and smaller than the one below it since the wad is slowing down at the rate of 10 m/s for each second. Similarly the successively higher points would become closer together as the wad slows down.
If you can't resist the temptation to come up with numerical results take a look at these: For 37 we have vi and a and displacement so the answer is
square root (2 a displ+ vi^2) = sqrt(2(-10)(-10)+0) = 14.1 m/s
For 40 we have delta t = [+sqrt (2(-10)(3) +100)-10]/-10 = .37 sec or there abouts.
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