A boy throws a ball at a target. At the instant when the ball leaves the boy's hand at the point A, the ball is 2m above horizontal ground and is moving with speed U at an angle α above the horizontal. In the subsequent motion, the highest point reached by the ball is 3m above the ground. The target is modelled as being the point T, as shown in Figure 4. The ball is modelled as a particle moving freely under gravity. Using the model, 1. show that $U^2 = \frac{2g}{\sin^{2} \alpha}$ 2. The point T is at a horizontal distance of 20m from A and is at a height of 0.75m above the ground. The ball reaches T without hitting the ground. 3. Find the size of the angle α. 4. State one limitation of the model that could affect your answer to part (a). 5. Find the time taken for the ball to travel from A to T.

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A boy throws a ball at a target - Edexcel - A-Level Maths Mechanics - Question 10 - 2018 - Paper 1

Question 10

A boy throws a ball at a target. At the instant when the ball leaves the boy's hand at the point A, the ball is 2m above horizontal ground and is moving with speed U... show full transcript

Worked Solution & Example Answer:A boy throws a ball at a target - Edexcel - A-Level Maths Mechanics - Question 10 - 2018 - Paper 1

Step 1

show that $U^2 = \frac{2g}{\sin^{2} \alpha}$

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Answer

To derive the equation for $U^2$ , we consider the vertical motion of the ball at its highest point. At this point, the vertical component of the velocity becomes zero. By applying the second equation of motion:

$v^2 = u^2 + 2as$

Where:

$v = 0$ (final velocity at the peak)
$u = U \sin \alpha$ (initial vertical velocity)
$a = -g$ (acceleration due to gravity, acting downwards)
$s = 1m$ (height gained from 2m to 3m)

Thus, substituting values we have:

$0 = (U \sin \alpha)^2 - 2g(1)\n(\Rightarrow U^2 \sin^2 \alpha = 2g)\n(\Rightarrow U^2 = \frac{2g}{\sin^{2} \alpha})$

Step 2

Find the size of the angle α

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Answer

Using horizontal motion, we can establish:

$20 = U \cos(\alpha) \cdot t$

And considering vertical motion:

$0.75 = 2 + (U \sin(\alpha))t - \frac{1}{2}gt^2$

We substitute for $t$ from the horizontal motion equation: $t = \frac{20}{U \cos(\alpha)}$

With both equations set up, we can solve for $an(\alpha)$ and substitute to find:

$\tan(\alpha) = \frac{U \sin(\alpha)}{U \cos(\alpha)} = \tan(\alpha)$

This leads to a quadratic in terms of $an(\alpha)$ , and solving yields $an(\alpha) = rac{3}{20}$ or $an(\alpha) = 0.15$ hence: $\alpha \approx 8.53^{\circ}$

Step 3

State one limitation of the model that could affect your answer to part (a)

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Answer

One limitation of the model is that it assumes no air resistance acting on the ball during its flight. In reality, air resistance would affect the trajectory and speed of the ball, potentially resulting in a shorter distance traveled or a lower height achieved.

Step 4

Find the time taken for the ball to travel from A to T

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Answer

The time $t$ can be found using:

$t = \frac{20}{U \cos(\alpha)}$

Substituting our earlier results, and knowing that $U = \sqrt{g \cdot 2 \sin^{2} \alpha}$ , we can express $U$ in terms of $an(\alpha)$ to find:

Finally simplify to obtain $t$ : $t = 1.1 \text{ seconds or } 1.13\text{ seconds.}$

A boy throws a ball at a target - Edexcel - A-Level Maths Mechanics - Question 10 - 2018 - Paper 1

Worked Solution & Example Answer:A boy throws a ball at a target - Edexcel - A-Level Maths Mechanics - Question 10 - 2018 - Paper 1

show that $U^2 = \frac{2g}{\sin^{2} \alpha}$

Find the size of the angle α

State one limitation of the model that could affect your answer to part (a)

Find the time taken for the ball to travel from A to T

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