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 $T$, we can use the formula for time taken to reach the ground:

$T = \frac{u + v}{g}$

where:

• $u = 22.4$ m/s,
• $v = - ext{when hitting ground}$.

First, find the total height from the point of projection:

$h = 1.5 + 25.6 = 27.1$ m.

Now, using the equation:

$s = ut + \frac{1}{2} at^2$

Setting $s = -27.1$ m (downwards, hence negative), we can solve for $T$.

After performing calculations, we find: $T \approx 4.64 \text{ seconds}$

The resistive force $F$ can be calculated using the work-energy principle. The work done by the resistive force is equal to the kinetic energy of the ball just before it hits the ground:

The ball sinks 2.5 cm = 0.025 m.

Using: $ext{Work} = F \times d$ $F \times 0.025 = \frac{1}{2} m v^2$

• where $m = 0.6$ kg and using the previously found $v$ just before hitting the ground: $v = 22.4 + 9.8T \implies v = 22.4 m/s (directly at impact)$ Calculating: $F = \frac{0.5 \times 0.6 \times (22.4)^2}{0.025} = 6390 \text{ N} (to 3 significant figures)$

One physical factor that could be considered is air resistance. In reality, as the ball moves upwards and downwards, it would encounter air drag, which would require a modified model to accurately represent the motion.