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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 ... 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
Answer
To find the time taken for the stone to travel from O to A, we can use the vertical motion equation:
[ s = ut + \frac{1}{2} a t^2 ]
Where:
- s = -70 m (the stone moves downwards)
- u = 65 \sin(a)
- a = -10\ ms⁻² (acceleration due to gravity)
From ( \tan a = \frac{5}{12} ), we can calculate ( \sin a ) and ( \cos a ):
[ \sin a = \frac{5}{13} \quad \text{and} \quad \cos a = \frac{12}{13} ]
Substituting into the vertical motion equation:
[ -70 = 65 \cdot \frac{5}{13} t - \frac{1}{2} \cdot 10 t^2 ] [ -70 = 25t - 5t^2 ] [ 5t^2 - 25t - 70 = 0 ]
Using the quadratic formula ( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ):
[ t = \frac{25 \pm \sqrt{(-25)^2 - 4 \cdot 5 \cdot (-70)}}{2 \cdot 5} ] [ t = \frac{25 \pm \sqrt{625 + 1400}}{10} = \frac{25 \pm \sqrt{2025}}{10} = \frac{25 \pm 45}{10} ]
Calculating gives us: [ t = 7 \text{ seconds (taking the positive root)} ]
Step 2
find the speed of the stone at the instant just before it hits the ground at A
Answer
To find the speed of the stone just before it hits the ground, we need to calculate both the horizontal and vertical components of the velocity.
The horizontal component of the velocity is given by: [ V_x = 65 \cos(a) ] Using ( \cos a = \frac{12}{13} ): [ V_x = 65 \cdot \frac{12}{13} = 60 \text{ ms}^{-1} ]
The vertical component can be calculated using: [ V_y = u_y + a t ] Where:
- ( u_y = 65 \sin(a) = 65 \cdot \frac{5}{13} = 25 \text{ ms}^{-1} )
- a = -10 ms⁻²
- t = 7 s
Substituting gives: [ V_y = 25 - 10 \cdot 7 = 25 - 70 = -45 \text{ ms}^{-1} ]
The overall speed of the stone just before it hits the ground will be: [ V = \sqrt{V_x^2 + V_y^2} ] [ V = \sqrt{60^2 + (-45)^2} = \sqrt{3600 + 2025} = \sqrt{5625} = 75 \text{ ms}^{-1} ]
Step 3
State one limitation of the model that could affect the reliability of your answers
Answer
One limitation of the model is that it ignores air resistance. Air resistance can significantly affect the trajectory and speed of a projectile, especially at high velocities. This simplification may lead to inaccuracies in predicting the time of flight and impact speed.