Mr. Rogers' AP Physics C: Kinematics Problems
Coin Flip Problem: Albert flips a coin up in the air at an upward velocity of 5 m/s. He fails to catch it on the way down and it falls down an 8 m deep well. Draw a vs t, v vs t, and y vs t curves. Calculate the max height and max magnitude of velocity of the coin.

Solution

Define downward as negative.  Clearly, the acceleration is constant so the a vs.t, v vs. t, x vs. t are as shown at bottom right.

 First find the time to the top as follows: vtop = at + vo 0 = at + vo - at = vo t = vo / (-a) t = (5 m/s) / (-(-10 m/s2)) t = 0.5 s to reach the top Next find the displacement at the top x = 1/2at2 + vo t + xo = 1/2 (-10 m/s2) (0.5 s)2 + (5 m/s) (0.5 s) + 0 = - 1.25 + 2.5 y = 1.25 m at the top The max velocity occurs at the bottom. First find the time to reach the bottom: x = 1/2at2 + vo t + xo -8 m = 1/2 (-10 m/s2) t2 + (5 m/s) t + 0 For simplicity we'll drop the units and rearrange. 0 = 5t2 - 5t - 8 Using the quadratic t = (5 ± [ 52 - 4(5)(-8) ] 0.5 ) / 2(5) = (5 ± [ 25 + 160 ] 0.5 ) / 10 = (5 ± 13.6) / 10 = 1.86 or - 0.86 obviously it's t = 1.83 to reach bottom Finally we find the final velocity vbot. = at + vo = (-10 m/s2)(1.86 s) + 5 m/s = -18.6 + 5 vbot. = - 13.6 m/s at the bottom

Conclusion and Significance

The final speed of the coin is greater than the starting speed because it falls to a position below the starting position. If it fell back to the same position, the speed would be identical to the starting speed.

Metacognition Questions:

Do the number of unknowns match the number of equations? Most problems in physics including the type illustrated above are solved by simply writing enough equations so that the number of unknowns equals the number of equations. The equations can then be solved simultaneously.

Is the acceleration constant? With constant acceleration there are essentially just two equations that can be written:
 v = at + vo x = 1/2at2 + vo t + xo

However, these equations are only valid for constant acceleration.