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AQA GCSE · Question 07.8 · Forces
The child steps off the diving board and falls into the swimming pool. The initial velocity of the child is 0 m/s.
<br>
acceleration due to gravity = 9.8 m/s²
<br><br>
Calculate the final velocity when the child has fallen a distance of 0.95 m through the air.
<br>
Give your answer to 2 significant figures.
<br>
Use the Physics Equations Sheet.
The child steps off the diving board and falls into the swimming pool. The initial velocity of the child is 0 m/s.
<br>
acceleration due to gravity = 9.8 m/s²
<br><br>
Calculate the final velocity when the child has fallen a distance of 0.95 m through the air.
<br>
Give your answer to 2 significant figures.
<br>
Use the Physics Equations Sheet.
How to approach this question
1. Look at the Physics Equations Sheet for an equation that links final velocity (v), initial velocity (u), acceleration (a), and distance (s). The correct equation is v² - u² = 2as.
2. Identify the values for u, a, and s from the question.
3. Substitute these values into the equation.
4. Solve the equation for v².
5. Take the square root to find v.
6. Round your final answer to 2 significant figures as requested.
Full Answer
v² - u² = 2as
v² - 0² = 2 × 9.8 × 0.95
v² = 18.62
v = √18.62
v = 4.315... m/s
v = 4.3 m/s (to 2 s.f.)
We are given:
Initial velocity (u) = 0 m/s
Acceleration (a) = 9.8 m/s²
Distance (s) = 0.95 m
We need to find the final velocity (v).
The appropriate equation of motion from the Physics Equations Sheet is:
(final velocity)² - (initial velocity)² = 2 × acceleration × distance
v² - u² = 2as
Substitute the known values:
v² - 0² = 2 × 9.8 × 0.95
v² = 18.62
Now, find v by taking the square root of both sides:
v = √18.62
v = 4.315089... m/s
The question asks for the answer to 2 significant figures.
v ≈ 4.3 m/s
Common mistakes
✗ Choosing the wrong equation of motion.\n✗ Forgetting to take the square root at the end.\n✗ Rounding incorrectly to 2 significant figures.
Practice the full AQA GCSE Physics Foundation Tier Paper 2
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