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, the total time taken to reach the ground, we can use the equation of motion:

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

Here, the total displacement s is the distance from point A to the ground:

• s = 1.5 m (initial height) + h (height traveled upwards) - 1.5 m.

Let’s solve for T:

$-1.5 = 22.4T - \frac{1}{2}(9.8)T^2$

Re-arranging gives:

$4.9T^2 - 22.4T - 1.5 = 0$

Using the quadratic formula, we find:

$T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

With a = 4.9, b = -22.4, and c = -1.5, the solution yields T to be approximately: $T \approx 4.64$ seconds (after calculation)

To calculate the resistive force F, we use the formula:

$F = ma$

where m is the mass of the ball (0.6 kg) and a is the deceleration when the ball comes to rest after sinking into the ground. Given that the ball sinks 2.5 cm (0.025 m), we can find a using:

$v^2 = u^2 + 2as$

Here, final velocity = 0, initial velocity = speed just before sinking, and distance s = 0.025 m. Using previous calculations:

$v = 22.4 - 9.8 \times 4.64$

Calculating F: $F = 0.6 \times (-1064.5) \approx -638.7 ext{ N} ext{ (to 3 significant figures)}$

One physical factor to consider is air resistance. The assumption of a constant upward speed does not account for the deceleration due to drag, which varies with the square of the speed and thus alters the motion of the ball significantly.