Two students, Student 1 and Student 2 are on a bridge. They both throw identical rocks into the river below. Student 1 throws a rock straight up at 15.00 m/s, whilst Student 2 throws an identical rock straight down at 15.00 m/s.The rock thrown by Student 2 takes 6.0 seconds to reach the river below. Ignore air resistance. What is the maximum height (m) reached by the rock thrown by Student 1? Give your answer correct to one decimal places.

Question

BlackTom AI · Accepted Answer

The problem involves two rocks thrown from the same height with equal speed but in opposite directions, and asks for the maximum height reached by the rock thrown upward.

First, analyze Student 2's rock to determine the initial height from which both rocks were launched. Treat downward as positive and use s = v0 t + (1/2) a t^2.
- v0 (downward) = 15 m/s, t = 6.0 s, a = g = 9.8 m/s^2.
- Distance to river: H = (15)(6) + 0.5(9.8)(6^2) = 90 + 0.5(9.8)(36) = 90 + 176.4 = 266.4 m.
So the launch height above the river is 266.4 m.

Next, focus on Student 1's rock, which is thrown upward with the same speed 15 m/s. To find the maximum height reached above the launch point, use the kinematic relation v^2 = u^2 + 2 a s, with final velocity v = 0 at the top, u = 15 m/s (upward), and a = -g = -9.8 m/s^2.
- 0 = (15)^2 + 2(-9.8)s
- Solve for s: 0 = 225 - 19.6 s  =>  s = 225 / 19.6 ≈ 11.45 m.
Thus the maximum height above the launch point is about 11.45 m, which rounds to 11.5 m to one decimal place.

In summary, the upward-throw rock rises to approximately 11.5 m above the launch point, while the launch height above the river is 266.4 m (from Student 2’s motion), but the question asks for the maximum height of Student 1's rock above the launch point, yielding about 11.5 m.

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