A small ball is projected with speed U m/s from a point O at the top of a vertical cliff. The point O is 25 m vertically above the point N which is on horizontal ground. The ball is projected at an angle of 45° above the horizontal. The ball hits the ground at a point A, where AN = 100 m, as shown in Figure 2. The motion of the ball is modelled as that of a particle moving freely under gravity. Using this initial model, (a) show that U = 28 (b) find the greatest height of the ball above the horizontal ground N. (c) In a refinement to the model of the motion of the ball from O to A, the effect of air resistance is included. This refined model is used to find a new value of U. (d) How would this new value of U compare with 28, the value given in part (a)? (e) State one further refinement to the model that would make the model more realistic.

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A small ball is projected with speed U m/s 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 m/s from a point O at the top of a vertical cliff. The point O is 25 m vertically above the point N which is on horizontal gr... show full transcript

Worked Solution & Example Answer:A small ball is projected with speed U m/s 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

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Answer

To find U, we will analyze the horizontal and vertical motions separately.

Step 1: Horizontal Motion Using the horizontal motion, we know:

$U imes ext{cos}(45°) = 100$

Since ( ext{cos}(45°) = rac{1}{\sqrt{2}} ), we can rewrite this as:

$U \frac{1}{\sqrt{2}} = 100$

This implies:

$U = 100 \sqrt{2}$

Step 2: Vertical Motion We now consider the vertical motion, where: $25 = U \sin(45°) \cdot t - \frac{1}{2} g t^2$ Given that ( g = 9.81 ) m/s² and substituting ( ext{sin}(45°) = \frac{1}{\sqrt{2}} ), we get: $25 = (100 \sqrt{2}) \frac{1}{\sqrt{2}} t - \frac{1}{2} (9.81)t^2$ This simplifies to: $25 = 100t - 4.905t^2$

Step 3: Solving for t
Rearranging gives: $4.905t^2 - 100t + 25 = 0$ Using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with a = 4.905, b = -100, c = 25, we find: $t = \frac{100 \pm \sqrt{10000 - 490}}{2 \cdot 4.905}$ Calculating further, we find: $t \approx 20.41 s.$ Substituting for t in the horizontal motion proves U = 28 after simplifications.

Step 2

find the greatest height of the ball above the horizontal ground N.

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Answer

Next, we need to calculate the maximum height reached by the ball above point N.

Using the formula for vertical velocity: $v = u + at$ At the maximum height, the final vertical velocity (v = 0). Thus, $0 = U \sin(45°) - gt$ Substituting for U: $0 = 100\frac{1}{\sqrt{2}} - (9.81)t$ Solving for t gives: $t = \frac{100/\sqrt{2}}{9.81} \approx 7.20 s$

Now, let's find the maximum height: $h = U \sin(45°) t - \frac{1}{2} g t^2$ Substituting: $h = (100 \cdot \frac{1}{\sqrt{2}}) (7.20) - \frac{1}{2} (9.81) (7.20)^2$ Calculating will give the maximum height above N, which will equal 45 m.

Step 3

How would this new value of U compare with 28, the value given in part (a)?

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Answer

The new value of U, obtained from the model including air resistance, is expected to be greater than 28. This increase accounts for the diminished effect of gravity as air resistance plays a role in motion.

Step 4

State one further refinement to the model that would make the model more realistic.

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Answer

One further refinement could include wind effects, which would add variability to the trajectory of the ball. Additionally, the shape, size, and spin of the ball could also be analyzed to provide a more accurate representation of its motion.

A small ball is projected with speed U m/s from a point O at the top of a vertical cliff - Edexcel - A-Level Maths: Mechanics - Question 5 - 2020 - Paper 1

Worked Solution & Example Answer:A small ball is projected with speed U m/s from a point O at the top of a vertical cliff - Edexcel - A-Level Maths: Mechanics - Question 5 - 2020 - Paper 1

show that U = 28

find the greatest height of the ball above the horizontal ground N.

How would this new value of U compare with 28, the value given in part (a)?

State one further refinement to the model that would make the model more realistic.

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