A stone is projected vertically upwards from a point A with speed u ms−1. After projection the stone moves freely under gravity until it returns to A. The time between the instant that the stone is projected and the instant that it returns to A is rac{3}{4} seconds. Modelling the stone as a particle, (a) show that $u = 17 rac{1}{2}$, (b) find the greatest height above A reached by the stone, (c) find the length of time for which the stone is at least $6 rac{3}{5}$ m above A.

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

Question 5

A stone is projected vertically upwards from a point A with speed u ms−1. After projection the stone moves freely under gravity until it returns to A. The time betwe... show full transcript

Worked Solution & Example Answer:A stone is projected vertically upwards from a point A with speed u ms−1 - Edexcel - A-Level Maths Mechanics - Question 5 - 2012 - Paper 1

Step 1

show that $u = 17 rac{1}{2}$

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Answer

To show that the initial speed $u$ is $17 rac{1}{2}$ , we use the equation of motion which states:

$v = u + at$

At the highest point, the final velocity $v = 0$ m/s. Also, the acceleration due to gravity $a = -g = -9.8$ m/s².

The total time for the stone to travel up and down is $T = rac{3}{4}$ seconds, hence the time to reach the highest point is:

$t = rac{T}{2} = rac{3}{8} ext{ seconds}$

Substituting the values into the equation:

$0 = u - 9.8 imes rac{3}{8}$

Thus,

$u = 9.8 imes rac{3}{8} = 17.25$

Therefore, we have shown that $u = 17 rac{1}{2}$ .

Step 2

find the greatest height above A reached by the stone

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Answer

To find the greatest height reached, we utilize the following equation of motion:

$s = ut + rac{1}{2}at^2$

For the upward motion,

$u = 17.25$ m/s
$a = -9.8$ m/s²
$t = rac{3}{8}$ s

Substituting into the equation:

$s = 17.25 imes rac{3}{8} + rac{1}{2} imes (-9.8) imes igg( rac{3}{8}igg)^2$

Calculating,

$s = 17.25 imes 0.375 - 0.5 imes 9.8 imes rac{9}{64}$

Thus:

$s = 6.46875 - 0.5 imes 9.8 imes 0.140625 \s = 6.46875 - 0.68625$

Finally,

$s ext{ (greatest height)} ext{ is approximately } 5.7825 ext{ m above A}$

Step 3

find the length of time for which the stone is at least $6 rac{3}{5}$ m above A

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Answer

First, convert $6 rac{3}{5}$ m to an improper fraction:

$6 rac{3}{5} = rac{33}{5}$

Using the equation of motion:

$s = ut + rac{1}{2} at^2$

Set the equation for when the height $s$ is at least $6.6$ m:

$17.25t - 4.9t^2 ext{ must satisfy } \ 17.25t - 4.9t^2 ext{ ≥ } 6.6$

This gives:

$-4.9t^2 + 17.25t - 6.6 = 0$

Using the quadratic formula:

$t = rac{-b ext{ ± } ext{√}(b^2 - 4ac)}{2a}$

Here, $a = -4.9$ , $b = 17.25$ , and $c = -6.6$ , so substituting values gives:

$t = rac{-17.25 ext{ ± } ext{√}(17.25^2 - 4 imes (-4.9) imes (-6.6))}{2 imes -4.9}$

Solving this yields the times for which the height is above $6 rac{3}{5}$ m.

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

Worked Solution & Example Answer:A stone is projected vertically upwards from a point A with speed u ms−1 - Edexcel - A-Level Maths Mechanics - Question 5 - 2012 - Paper 1

show that $u = 17 rac{1}{2}$

find the greatest height above A reached by the stone

find the length of time for which the stone is at least $6 rac{3}{5}$ m above A

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