A ball is thrown straight up at an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. Using g = 9.80 m/s^2, how much additional time passes before it passes that same branch again on the way down?
A0.574 s
B0.956 s
C1.53 s
D1.91 s
Answer & Solution
Correct answer: D. 1.91 s
1. Solve 7.00 = 15.0t - 4.90t^2, or 4.90t^2 - 15.0t + 7.00 = 0.
2. The quadratic formula gives t = (15.0 +/- square root of 87.8) / 9.80.
3. The two roots are t_up = 0.574 s and t_down = 2.487 s.
4. The additional time between the two passes is 2.487 s - 0.574 s = 1.91 s.
_Source: OpenStax Physics (CC BY 4.0), Ch 3 "Acceleration", section 3.2 Representing Acceleration with Equations and Graphs_
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