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 11.94 ms-1, whilst Student 2 throws an identical rock straight down at 11.94 ms-1.The rock thrown by Student 2 takes 2.24 seconds to reach the river below. Ignoring air resistance, what distance( in m ) does the rock thrown by Student 2 travel when it reaches the river below? Give your answer to 2 decimal places without units.

Question

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 11.94 ms-1, whilst Student 2 throws an identical rock straight down at 11.94 ms-1.The rock thrown by Student 2 takes 2.24 seconds to reach the river below. Ignoring air resistance, what distance( in m ) does the rock thrown by Student 2 travel when it reaches the river below? Give your answer to 2 decimal places without units.

BlackTom AI · Accepted Answer

Use the kinematic equation for displacement with constant acceleration: s = u t + (1/2) a t^2. Here we want the distance traveled (magnitude of downward displacement), the initial velocity u is 11.94 m/s downward, the time t = 2.24 s, and acceleration a = g = 9.81 m/s^2 downward.

Step-by-step calculation:
1. Compute u t: 11.94 × 2.24 = 26.7456.
2. Compute t^2: 2.24^2 = 5.0176.
3. Compute (1/2) g t^2: 0.5 × 9.81 × 5.0176 = 4.905 × 5.0176 = 24.611328.
4. Sum the two contributions: s = 26.7456 + 24.611328 = 51.356928.

Rounding to two decimal places gives 51.36 (without units), matching the provided answer.

Why this is correct: the formula s = u t + 1/2 a t^2 gives the total displacement under constant acceleration. Because both initial velocity and gravity act downward, their contributions add. The calculation uses g = 9.81 m/s^2; using g = 9.80 would give a slightly different value (≈51.33), which is a common small variation, but the intended value here uses 9.81.

Common incorrect approaches and why they fail:
- Forgetting the (1/2) a t^2 term (i.e., using only u t) would omit the contribution of gravity and undercount the distance.
- Using the wrong sign conventions could lead to a negative displacement; distance requires the positive magnitude.
- Using g = 9.80 instead of 9.81 yields 51.33, which does not match the official answer.

Final answer confirmed: 51.36.

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