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

Using the kinematic equation to find the height at time t:

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

Here, s = 19 m above point A, u = 21 m/s, and a = -9.8 m/s².

Substituting the known values:

$19 = 21t - \frac{1}{2} (9.8)t^2$

Rearranging gives us:

$4.9t^2 - 21t + 19 = 0$

Applying the quadratic formula:

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

Where a = 4.9, b = -21, and c = 19:

$t = \frac{21 \pm \sqrt{(-21)^2 - 4(4.9)(19)}}{2 imes 4.9}$

Calculating the discriminant:

$t = \frac{21 \pm \sqrt{441 - 372.4}}{9.8}$

$t = \frac{21 \pm \sqrt{68.6}}{9.8}$

This gives two possible values for t:

$t \approx 2.99 ext{ or } 1.30 ext{ seconds}$

Using the work-energy principle:

The work done on particle P when it sinks is equal to the change in kinetic energy:

Let d be the distance P sinks.

Total weight force acting downwards:

$F = mg + resistive \ force = 4g + 5000$

Where g = 9.8 m/s²:

Total force:

$F = 4 \times 9.8 + 5000 = 5000 + 39.2 = 5039.2 ext{ N}$

Using Newton's second law:

rac{1}{2}mv^2 = Fd

Where initial kinetic energy before sinking is:

$\frac{1}{2} \times 4 \times 28^2 = 1568 ext{ J}$

So,

$1568 = 5039.2d$

Thus, solving for d:

$d = \frac{1568}{5039.2} \approx 0.312 ext{ meters}$

Or approximately 0.31 m.