A firework rocket starts from rest at ground level and moves vertically. In the first 3 s of its motion, the rocket rises 27 m. The rocket is modelled as a particle moving with constant acceleration a m s². Find a) the value of a, b) the speed of the rocket 3 s after it has left the ground. After 3 s, the rocket burns out. The motion of the rocket is now modelled as that of a particle moving freely under gravity. c) Find the height of the rocket above the ground 5 s after it has left the ground.

A-Level Edexcel Maths Mechanics Constant Acceleration - 1D: A firework rocket starts from re

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

Question 2

A firework rocket starts from rest at ground level and moves vertically. In the first 3 s of its motion, the rocket rises 27 m. The rocket is modelled as a particle ... show full transcript

Worked Solution & Example Answer:A firework rocket starts from rest at ground level and moves vertically - Edexcel - A-Level Maths Mechanics - Question 2 - 2008 - Paper 1

Step 1

the value of a

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Answer

To find the acceleration $a$ , we can use the first equation of motion: $s = ut + \frac{1}{2} a t^2$ where:

$s = 27$ m (the distance travelled),
$u = 0$ m/s (initial speed, since it starts from rest),
$t = 3$ s (time).

Substituting the known values:

$27 = 0 + \frac{1}{2} a (3)^2$

This simplifies to: $27 = \frac{1}{2} a \cdot 9$

Multiplying through by 2: $54 = 9a$

Dividing by 9 gives: $a = 6 ext{ m/s}^2$ .

Step 2

the speed of the rocket 3 s after it has left the ground

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After 3 seconds, we can use the formula for the final velocity:
v = u + at. Where:

$u = 0$ m/s (initial speed),
$a = 6$ m/s² (as calculated),
$t = 3$ s.

Substituting the values:
v = 0 + 6 \times 3 = 18 ext{ m/s}.

Step 3

Find the height of the rocket above the ground 5 s after it has left the ground

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Answer

For time $t$ from 3 to 5 seconds, the rocket is in free fall. The height $s$ can be modeled using:
s = ut + \frac{1}{2} a t^2 where:

$u = 18$ m/s (velocity just after 3 seconds),
$a = -9.8$ m/s² (acceleration due to gravity),
$t = 2$ s (from 3 s to 5 s).

Substituting the values:
s = 18 \times 2 - \frac{1}{2}(9.8)(2^2)
= 36 - \frac{1}{2}(9.8)(4)
= 36 - 19.6 = 16.4 ext{ m}.

Now adding the initial height gained in the first 3 seconds:
Total height = initial height + height while falling = 27 + 16.4 = 43.4 ext{ m}.

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

Worked Solution & Example Answer:A firework rocket starts from rest at ground level and moves vertically - Edexcel - A-Level Maths Mechanics - Question 2 - 2008 - Paper 1

the value of a

the speed of the rocket 3 s after it has left the ground

Find the height of the rocket above the ground 5 s after it has left the ground

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