A small stone is projected with speed 65 ms⁻¹ from a point O at the top of a vertical cliff. Point O is 70 m vertically above the point N. Point N is on horizontal ground. The stone is projected at an angle a above the horizontal, where tan a = \frac{5}{12}. The stone hits the ground at the point A, as shown in Figure 3. The stone is modelled as a particle moving freely under gravity. The acceleration due to gravity is modelled as having magnitude 10 ms⁻². Using the model, a) Find the time taken for the stone to travel from O to A, b) Find the speed of the stone at the instant just before it hits the ground at A. c) State one limitation of the model that could affect the reliability of your answers.

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A small stone is projected with speed 65 ms⁻¹ from a point O at the top of a vertical cliff - Edexcel - A-Level Maths Mechanics - Question 4 - 2021 - Paper 1

Question 4

A small stone is projected with speed 65 ms⁻¹ from a point O at the top of a vertical cliff. Point O is 70 m vertically above the point N. Point N is on horizontal g... show full transcript

Worked Solution & Example Answer:A small stone is projected with speed 65 ms⁻¹ from a point O at the top of a vertical cliff - Edexcel - A-Level Maths Mechanics - Question 4 - 2021 - Paper 1

Step 1

Find the time taken for the stone to travel from O to A

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Answer

To find the time taken, we use the vertical motion of the stone. The height from point O to N is 70 m, and the vertical component of velocity can be expressed as:

$y = v_{0y} t - \frac{1}{2} g t^2$

Given that $v_{0y} = 65 \sin a$ , where $\tan a = \frac{5}{12}$ , we can find $\sin a$ :

$\sin a = \frac{5}{13} \quad \text{and} \quad \cos a = \frac{12}{13}$

Thus, the vertical component is:

$v_{0y} = 65 \times \frac{5}{13} = 25 \text{ ms}^{-1}$

Substituting into the equation for vertical motion:

$-70 = 25t - \frac{1}{2} \times 10 t^2$

This simplifies to:

$5t^2 - 25t - 70 = 0$

Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , we find:

$t = \frac{25 \pm \sqrt{(-25)^2 - 4 \times 5 \times (-70)}}{2 \times 5} = \frac{25 \pm \sqrt{625 + 1400}}{10} = \frac{25 \pm \sqrt{2025}}{10}$

Calculating gives:

$t = \frac{25 + 45}{10} = 7 ext{ s}$

Therefore, the time taken for the stone to travel from O to A is 7 seconds.

Step 2

Find the speed of the stone at the instant just before it hits the ground at A

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Answer

To find the speed of the stone just before it hits the ground, we need to calculate both the horizontal ( $v_{x}$ ) and vertical ( $v_{y}$ ) components of the velocity at point A.

The horizontal velocity component remains constant and can be calculated as:

$v_{x} = 65 \cos a = 65 \times \frac{12}{13} = 60 \text{ ms}^{-1}$

For the vertical component at A, we use:

$v_{y} = v_{0y} - gt$

Substituting the values:

$v_{y} = 25 - 10 \times 7 = 25 - 70 = -45 \text{ ms}^{-1}$

Thus, the speed just before impact can be calculated using the Pythagorean theorem:

$v = \sqrt{v_{x}^2 + v_{y}^2} = \sqrt{(60)^2 + (-45)^2} = \sqrt{3600 + 2025} = \sqrt{5625} = 75 \text{ ms}^{-1}$

Therefore, the speed of the stone just before it hits the ground at A is 75 ms⁻¹.

Step 3

State one limitation of the model that could affect the reliability of your answers

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Answer

One limitation of the model is that it ignores air resistance, which can significantly affect the motion of the stone, especially at high speeds and over longer distances. This simplification means that our calculations of time and speed could overestimate the actual values.

A small stone is projected with speed 65 ms⁻¹ from a point O at the top of a vertical cliff - Edexcel - A-Level Maths Mechanics - Question 4 - 2021 - Paper 1

Worked Solution & Example Answer:A small stone is projected with speed 65 ms⁻¹ from a point O at the top of a vertical cliff - Edexcel - A-Level Maths Mechanics - Question 4 - 2021 - Paper 1

Find the time taken for the stone to travel from O to A

Find the speed of the stone at the instant just before it hits the ground at A

State one limitation of the model that could affect the reliability of your answers

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