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 $ heta$ above the horizontal, where $ an heta = rac{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 for the stone to travel from O to A, we can use the vertical motion equations.

The vertical distance fallen is 70 m, and the initial vertical component of velocity ( $v_{0y}$ ) can be determined by:

$v_{0y} = 65 imes rac{5}{13} \ ext{ (since } an heta = rac{5}{12} ext{ and the hypotenuse is 65)}$

Thus,

$v_{0y} = 65 imes rac{5}{13} = 25 ext{ ms}^{-1}$

Using the equation of motion for vertical displacement:

$s = v_{0y} t - rac{1}{2} g t^2$

we have:

$-70 = 25t - rac{1}{2} imes 10 imes t^2$

which simplifies to:

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

Using the quadratic formula,

$t = rac{-b imes rac{1}{5}}{2ac}$

we substitute,

$t = rac{25 imes rac{1}{5} imes ± rac{ ext{sqrt{(−25)^2 − 4 imes (5) × (−70)}}}{ ext{2 × (5)} = 7 ext{ seconds}}$

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 will find both horizontal and vertical components at point A.

Horizontal component: The horizontal component of the velocity is constant:

$v_{x} = 65 imes rac{12}{13} = 60 ext{ ms}^{-1}$
Vertical component: At point A, the vertical component can be evaluated using:

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

Substituting values, we get:

$v_{y} = 25 - 10 imes 7 = -45 ext{ ms}^{-1}$

Now we can determine the overall speed just before hitting the ground using Pythagorean theorem:

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

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 actual trajectory and speed of the stone during its flight.

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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