A particle P is projected vertically upwards from a point A with speed u ms⁻¹ - Edexcel - A-Level Maths Mechanics - Question 5 - 2012 - Paper 1

At time t seconds after projection, P is 19 m above A, so the total height h above ground is 19 + 17.5 = 36.5 m. Using the equation:

$h = ut + \frac{1}{2}(-9.8)t^2$

Substituting h and rearranging:

$36.5 = 21t - 4.9t^2$

This can be rearranged to:

$4.9t^2 - 21t + 36.5 = 0$

Now, using the quadratic formula:

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

where a = 4.9, b = -21, c = 36.5:

$t = \frac{21 \pm \sqrt{(-21)^2 - 4(4.9)(36.5)}}{2(4.9)}$

This results in:

$t = \frac{21 \pm \sqrt{441 - 714.8}}{9.8}$

Calculate to find potential t values giving t ≈ 2.99 or t ≈ 1.30.

When P reaches the ground, its upward motion stops and it begins to sink, subject to a resistive force. Using Newton's Second Law:

$F_{net} = ma$

Here, the net force acting on P just after reaching the ground is:

$F_{net} = mg - 5000$

Where m = 4 kg and g = 9.8 m/s²:

$F_{net} = (4)(9.8) - 5000 = 39.2 - 5000 = -4960.8 ext{ N}$

Set the net force equal to mass times acceleration:

$-4960.8 = 4a\implies a = -1240.2 m/s^2$

Using the kinematic equation again:

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

Where u = 0 (initial speed when sinking starts), v need to be found. Rewriting gives:

$0 = 5000s + 2(-1240.2)s$

Solving this, we find s ≈ 0.32 m (or 32 cm) is the depth that P sinks into the ground before coming to rest.