A ball is dropped from rest. It falls 1.8 m to the ground.
The acceleration of the ball is 10 m/s2. Ignore air resistance.
Use the equation: (final velocity)2−(initial velocity)2=2×acceleration×distance(\text{final velocity})^2 - (\text{initial velocity})^2 = 2 \times \text{acceleration} \times \text{distance}(final velocity)2−(initial velocity)2=2×acceleration×distance
State the initial velocity of the ball.
Calculate the velocity of the ball just before it hits the ground.
The ball is now dropped from rest from a height of 3.6 m.
A student says:
'The height has doubled, so the ball will hit the ground at twice the velocity.'
Explain why the student is wrong.
43 exam-style questions on OCR GCSE Physics 2.1 Motion, covering 2.1.1 Measuring distance and time, 2.1.2 Measuring speed, 2.1.3 Ratios, units and rates in motion, 2.1.4 Scalars and vectors in motion, 2.1.5 Distance-time and velocity-time graphs, 2.1.6 Area under a velocity-time graph, 2.1.7 Average speed for non-uniform motion, and 2.1.8 Uniform motion and uniform acceleration. Each one has a worked solution and a mark scheme showing where the marks go.