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 α above the horizontal, where tan α = 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 horizont... 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 will first analyze the vertical motion.
Using the equation of motion for vertical displacement, we have the following:

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

$s = -70$ m (the stone travels downward),
$u = 65 \sin(\alpha)$ ms⁻¹ (initial vertical component),
$a = -10$ ms⁻² (acceleration due to gravity, acting downwards).

Using the relationship $\tan(\alpha) = \frac{5}{12}$ , we can find $\alpha$ :

Calculate $\sin(\alpha)$ $sin (α)$ and $\cos(\alpha)$ $cos (α)$ using the ratio.
- $\sin(\alpha) = \frac{5}{\sqrt{5^2 + 12^2}} = \frac{5}{13}$
- $\cos(\alpha) = \frac{12}{\sqrt{5^2 + 12^2}} = \frac{12}{13}$

Thus,

$u = 65 \cdot \frac{5}{13} = 25 ms^{-1}.$

Substituting into the vertical motion equation:

$-70 = 25t - \frac{1}{2} \cdot 10 t^2$ or distributing terms leads us to:

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

Using the quadratic formula, we find:

= \frac{25 \pm \sqrt{625 + 1400}}{10} = \frac{25 \pm \sqrt{2025}}{10} = \frac{25 \pm 45}{10}.$$ Points yield: - $t = 7$ s (valid), - $t = -2$ s (not valid). Thus, 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 impact, we need to evaluate the horizontal and vertical components of velocity at point A.
The horizontal component remains constant:

$V_{x} = 65 \cos(\alpha) = 65 \cdot \frac{12}{13} = 60 ms^{-1}.$

For the vertical component, just before hitting the ground: Using the formula:

displacement and acceleration due to gravity gives us: $V_{y} = u + at = 25 + (-10)\cdot7 = 25 - 70 = -45 ms^{-1}.$

The magnitude of the total velocity is found using the Pythagorean theorem: $V = \sqrt{V_{x}^2 + V_{y}^2} = \sqrt{60^2 + (-45)^2} = \sqrt{3600 + 2025} = \sqrt{5625} = 75 ms^{-1}.$ Therefore, the speed of the stone at the instant 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 could slow down the stone as it travels through the air, thus affecting both the time of flight and the speed just before impact.

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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