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| Photo used with permission under Creative Commons license from Dennis Jarvis. |
Suppose a cannonball is fired vertically from ground-height at an initial velocity of 30 m/s. How long is the cannonball in flight before it crashes back to the ground? (Assume no air resistance in this problem.)The easy way to do this problem is calculate how long the cannonball takes to get to the peak of its trajectory:
v_f = v_i + a*t
0 = 30 + (-9.8) * t
t = 3.06 seconds
Since, by our symmetry argument, the cannonball spends an equal time going down as it does going up, we can double our result to find the overall time:
total time = 2 * (time going up)
total time = 2 * 3.06
total time = 6.12 seconds
While this is a quick and easy way to calculate the answer, does it make assumptions that are not true? Is the symmetry argument really valid in this case? Is there a way to determine the answer of 6.12 seconds of total flight time without using the symmetry argument?
In the comments, explain how it is possible to get t = 6.12 seconds without using the symmetry argument. What equations would you use and how do they work out?




