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

To find the time $T$ taken to reach the ground, we use the equation: $s = ut + \frac{1}{2}at^2$ Here, $s = -1.5$ m (the ball travels downward), $u = 22.4$ m/s, and $a = -9.8$ m/s². Substituting these values: $-1.5 = 22.4t - 4.9t^2$ Rearranging gives: $4.9t^2 - 22.4t - 1.5 = 0$ Using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ with $a = 4.9$, $b = -22.4$, and $c = -1.5$. We find: $T \approx 4.64 ext{ seconds (to 2 decimal places)}$

Using the work-energy principle, the potential energy lost equals the work done against the resistive force:

The height the ball sinks into the ground is: $d = 2.5 ext{ cm} = 0.025 ext{ m}$. The ball's weight is:\n$mg = 0.6 \text{ kg} \times 9.8 \text{ m/s}^2 = 5.88 ext{ N}.$\nThus, the work done against resistive force $F$: $F \cdot d = mg \cdot d$ Substituting:

One physical factor that could be considered is air resistance, which would affect the motion of the ball as it travels upwards and downwards.