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9709 P4 - Nov 2003 - Q2
3876
A stone is released from rest and falls freely under gravity. Find
the speed of the stone after 2 s,
the time taken for the stone to fall a distance of 45 m from its initial position,
the distance fallen by the stone from the instant when its speed is 30 m s-1 to the instant when its speed is 40 m s-1.
Solution
To find the speed after 2 s, use the formula for velocity under constant acceleration: \(v = u + gt\). Since the stone is released from rest, \(u = 0\). Thus, \(v = 0 + 10 \times 2 = 20 \text{ m s}^{-1}\).
To find the time taken to fall 45 m, use the formula \(s = \frac{1}{2}gt^2\). Solving \(45 = \frac{1}{2} \times 10 \times t^2\) gives \(t^2 = 9\), so \(t = 3 \text{ s}\).
To find the distance fallen from 30 m s-1 to 40 m s-1, use \(v^2 = u^2 + 2gs\). Substituting \(40^2 = 30^2 + 2 \times 10 \times s\), we get \(1600 = 900 + 20s\). Solving gives \(s = 35 \text{ m}\).
