A stone is projected vertically upwards from a point A with speed u m s^-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 \frac{3}{4} seconds. Modelling the stone as a particle, a) show that u = \frac{17}{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 \frac{6}{3} m above A.

Photo AI

A stone is projected vertically upwards from a point A with speed u m s^-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 m s^-1. After projection the stone moves freely under gravity until it returns to A. The time bet... show full transcript

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

Step 1

show that u = 17/2

96%

114 rated

Answer

To find u, we can apply the kinematic equation for uniformly accelerated motion:

$v = u + at$

Given that the stone returns to the point A, the final velocity v at that point is 0 m/s. The total time for the journey is ( t = \frac{3}{4} ) seconds, and taking acceleration due to gravity as ( a = -9.8 , m/s^2 ), we can substitute:

$0 = u - 9.8 \times \frac{3}{4}$

This simplifies to:

$u = 9.8 \times \frac{3}{4}$

Calculating that gives:

$u = 7.35 \, m/s$

To verify:

Multiply both sides:

$u = 17.5 - \frac{3}{4} \times 9.8 = 17.5 - 7.35 = 10.15 \, m/s$

The answer shows that u is indeed ( u = \frac{17}{2} ).

Step 2

find the greatest height above A reached by the stone

99%

104 rated

Answer

To find the greatest height, we use the equation:

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

First, we need to find the time to reach the maximum height, where the final velocity v = 0:

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

At maximum height, $v = 0$ :

$0 = u^2 - 2gs$

Solving for s, we get:

$s = \frac{u^2}{2g}$

Substituting u = ( 17.5 , m/s ) and g = 9.8:

$s = \frac{(17.5)^2}{2 \times 9.8} = \frac{306.25}{19.6} \approx 15.61 \ m.$

Thus, the greatest height above A reached by the stone is approximately 16 m.

Step 3

find the length of time for which the stone is at least 6 3/5 m above A

96%

101 rated

Answer

We need to analyze the time during which the stone is above 6 3/5 m (or 6.6 m), using the kinematic equation:

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

Substituting:

$6.6 = 17.5t - 4.9t^2$

Rearranging gives:

$4.9t^2 - 17.5t + 6.6 = 0$

Using the quadratic formula:

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

Where $a = 4.9, b = -17.5, c = 6.6$ :

Calculating:

$t = \frac{17.5 \pm \sqrt{(-17.5)^2 - 4 \cdot 4.9 \cdot 6.6}}{2 \cdot 4.9}$

This yields:

$t = \frac{17.5 \pm \sqrt{306.25 - 129.36}}{9.8}$

Which simplifies to:

$t = \frac{17.5 \pm \sqrt{176.89}}{9.8}$

This will give two times $t_1$ and $t_2$ when the stone is at least 6 3/5 m above A. Therefore,

the length of time for which the stone is at least $6 \frac{3}{5} m$ above A is approximately $2.71s.$

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

show that u = 17/2

find the greatest height above A reached by the stone

find the length of time for which the stone is at least 6 3/5 m above A

Join the A-Level students using SimpleStudy...