A ball is projected vertically upwards with a speed u m s⁻¹ from a point A which is 1.5 m above the ground - Edexcel - A-Level Maths Mechanics - Question 7 - 2003 - Paper 1

Using the kinematic equation: $s = ut + \frac{1}{2} a t^2$ where:

• s = -1.5 m (distance traveled to the ground)
• u = 22.4 m/s
• a = -9.8 m/s²

Setting up the equation: $-1.5 = 22.4t - 4.9t^2$ Rearranging gives: $4.9t^2 - 22.4t - 1.5 = 0$ Using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where ( a = 4.9, b = -22.4, c = -1.5 ) Calculating: $t = \frac{22.4 \pm \sqrt{(-22.4)^2 - 4 \times 4.9 \times (-1.5)}}{2 \times 4.9}$ Evaluating gives approximately: $t \approx 4.64 \, seconds$

The ball sinks 2.5 cm (or 0.025 m) into the ground. Using the work-energy principle: $\text{Work done} = \text{Force} \times \text{distance}$ The work done by the resistive force F should equal the change in kinetic energy when the ball comes to rest: $F \cdot 0.025 = \frac{1}{2} \cdot m \cdot v^2$ Substituting in the values:

• m = 0.6 kg, v = \sqrt{22.4^2 + 2 \times (-9.8) \times (-1.5)}$which yields v = 23.07 m/s (prior to sinking). Thus:$F \cdot 0.025 = \frac{1}{2} \cdot 0.6 \cdot (23.07)^2$Solving for F gives:$F = \frac{0.5 \cdot 0.6 \cdot 23.07^2}{0.025} \approx 639.0 \space N$$

Air resistance, which varies with the speed of the ball, affecting its upward and downward motion.