A firework rocket starts from rest at ground level and moves vertically - Edexcel - A-Level Maths Mechanics - Question 2 - 2008 - Paper 1

To find the speed $v$ of the rocket after 3 seconds, we can use the formula:

$v = u + at$

Here, $u = 0$, $a = 6 ext{ m/s}^2$, and $t = 3$ s:

$v = 0 + 6 \cdot 3 = 18 ext{ m/s}.$

After 3 seconds, the rocket continues to move under the influence of gravity. For the motion from $t = 3$ seconds to $t = 5$ seconds, the time $t = 2$ s. The height $s$ above the ground can be calculated using:

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

In this case, initial velocity $u = 18 ext{ m/s}$ (speed at t = 3 s), acceleration $a = -9.8 ext{ m/s}^2$ (downward due to gravity), and $t = 2$ s:

$s = 18 \cdot 2 + \frac{1}{2} \cdot (-9.8) \cdot (2^2)$

Calculating gives:

$s = 36 - \frac{1}{2} \cdot 9.8 \cdot 4$ $s = 36 - 19.6 = 16.4 ext{ m}.$

Thus, the total height above the ground after an additional 5 seconds is:

$\text{Total Height} = s + 27 = 16.4 + 27 = 43.4 ext{ m}.$