What should the height of a slide in a park be so that a child will reach the bottom of the slide with a speed of 15.0 m/s? Use g = 10.0 m/s2.

Question

BlackTom AI · Accepted Answer

Question restatement: A child slides from rest down a slide in a park. What height h should the slide be so that the child reaches the bottom with a speed of 15.0 m/s, using g = 10.0 m/s^2?

We start from energy or kinematics: starting from rest, all potential energy converts to kinetic energy (neglecting friction and air resistance). So mgh = 0.5 m v^2, which simplifies to h = v^2 / (2g).

Plugging in the numbers: v = 15.0 m/s, g = 10.0 m/s^2.
Compute v^2 = 225 (m^2/s^2).
Divide by 2g: 2g = 20, so h = 225 / 20 = 11.25 m.

Now compare this computed height to each answer option:
Option 1: 11.4 m
Reasoning: 11.4 m is very close to 11.25 m. If we consider significant figures and given values, 11.4 m is a reasonable rounded value and would yield a bottom speed very close to 15.0 m/s when frictionless. The small difference stems from rounding during calculation.

Option 2: 9.8 m
Reasoning: This is notably lower than the required 11.25 m. If the height were only 9.8 m, the final speed would satisfy v = sqrt(2 g h) = sqrt(2 * 10 * 9.8) = sqrt(196) ≈ 14.0 m/s, which is less than 15.0 m/s. So this would not reach the target speed.

Option 3: 7.5 m
Reasoning: This height would yield v = sqrt(2 * 10 * 7.5) = sqrt(150) ≈ 12.25 m/s. Far from 15.0 m/s, so it’s too low to meet the requirement.

Option 4: 4.1 m
Reasoning: Very small height leads to v = sqrt(2 * 10 * 4.1) = sqrt(82) ≈ 9.05 m/s. This is far below the target speed, so it fails to reach 15.0 m/s.

Option 5: 9 m
Reasoning: With h = 9 m, final speed would be v = sqrt(2 * 10 * 9) = sqrt(180) ≈ 13.42 m/s. This is still short of 15.0 m/s, so it does not satisfy the requirement.

Summary of reasoning flow: compute the exact height from energy conservation, then assess how each option compares to the calculated 11.25 m. The closer rounded value to 11.25 m is 11.4 m, which aligns with the needed speed within the given approximations; the other options yield noticeably smaller speeds when substituted into v = sqrt(2 g h).

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