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A small ball is projected with speed $U ext{ m s}^{-1}$ from a point $O$ at the top of a vertical cliff - Edexcel - A-Level Maths Mechanics - Question 5 - 2020 - Paper 1
Question 5
A small ball is projected with speed $U ext{ m s}^{-1}$ from a point $O$ at the top of a vertical cliff. The point $O$ is 25 m vertically above the point $N$ whic... show full transcript
Worked Solution & Example Answer:A small ball is projected with speed $U ext{ m s}^{-1}$ from a point $O$ at the top of a vertical cliff - Edexcel - A-Level Maths Mechanics - Question 5 - 2020 - Paper 1
Step 1
show that $U = 28$
Answer
To find , we can utilize both horizontal and vertical motion equations derived from the projectile's path.
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Horizontal Motion: The horizontal distance traveled is given by:
where . Thus,
U imes rac{1}{ ext{sqrt}(2)} imes t = 100 or
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Vertical Motion: The vertical distance covered is:
y = U imes ext{sin}(45^{ ext{°}}) imes t - rac{1}{2} g t^2 where (as it's downward), and . Thus,
-25 = U imes rac{1}{ ext{sqrt}(2)} imes t - rac{1}{2} (9.81) t^2. ag{2}
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Elimination and Solving for : By substituting into , we can solve for . Arranging yields:
Step 2
find the greatest height of the ball above the horizontal ground $N$
Answer
To find the greatest height, we need to calculate the maximum height from the point before the projectile begins descending. Since the vertical motion is considered:
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Vertical Motion Maximum Height Formula: The height can be found using:
h = U imes ext{sin}(45^{ ext{°}}) imes t - rac{1}{2} g t^2.
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Finding at Maximum Height: At maximum height, vertical velocity becomes zero:
Therefore,
t_{max} = rac{U imes rac{1}{ ext{sqrt}(2)}}{g}.
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Substituting and Solving for Height: By substituting the value of and solving for height, we yield:
h = U imes rac{1}{ ext{sqrt}(2)} imes rac{U}{g} - rac{1}{2} g imes rac{U^2}{g^2} = 45 ext{ m}.
Step 3
How would this new value of $U$ compare with 28?
Answer
The inclusion of air resistance in the model typically reduces the speed required to reach a given distance because some energy is dissipated due to drag forces acting against the motion of the ball. Thus:
- The new value of would likely be greater than 28, indicating that more thrust is needed to overcome the drag forces and achieve the height and distance achieved in the initial model.
Step 4
State one further refinement to the model that would make the model more realistic.
Answer
One refinement could be to include factors such as wind effects, as they can significantly influence the trajectory of the projectile. Additionally, incorporating the shape of the ball and how it interacts with the air could provide a more precise model.