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, (a) 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. (b) Find the size of the angle α. (9) (c) State one limitation of the model that could affect your answer to part (a). (1) (d) Find the time taken for the ball to travel from A to T. (3)

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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 this equation, we analyze the vertical motion of the ball at its highest point. At the peak, the vertical component of velocity is zero, so we can use the kinematic equation:

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

Setting the final velocity $v = 0$ , initial velocity $u = U \sin \alpha$ , acceleration $a = -g$ (downwards), and displacement $s = 3 - 2 = 1$ m (from 2m to 3m):

$0 = (U \sin \alpha)^2 - 2g(1)$

Rearranging gives:

$(U \sin \alpha)^2 = 2g$

Thus,

$U^2 \sin^2 \alpha = 2g,$

which leads us to:

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

Step 2

Find the size of the angle α.

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Answer

We apply the horizontal motion equation to find the angle α. The horizontal distance is given as 20m, and we know the horizontal velocity is $U \cos \alpha$ .

Using the equation:

$d = vt$

where $d = 20m$ and $t$ is the time taken. We also have:

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

Next, using the time of flight from part (d) and substituting it into the horizontal equation allows us to solve for $\tan \alpha$ :

$\tan \alpha = \frac{h}{d} = \frac{0.75 - 2}{20} = -\frac{1.25}{20},$

and we can deduce:

$\alpha = \tan^{-1}(-0.0625).$

Calculating gives:

$\alpha \approx 14^{\circ}.$

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 the assumption that there is no air resistance. In reality, air resistance would affect the motion of the ball, particularly at high speeds, causing it to travel a shorter distance and possibly altering the highest point reached.

Step 4

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

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Answer

To find the time taken, we can use the vertical motion equations. The initial vertical velocity is $U \sin \alpha$ , and the total displacement is from the height of 2m to 0.75m. We can use:

$s = ut + \frac{1}{2}at^2,$

where $s = -1.25 m$ , $u = U \sin \alpha$ , and $a = -g$ .

Substituting into the equation:

$-1.25 = (U \sin \alpha)t - \frac{1}{2}gt^2.$

This can be solved for time $t$ . With numerical values for $U$ , $g$ , and the angle $\alpha$ , we will find:

$$t = \frac{5}{\sqrt{2g}} \approx 1.1 $or$ 1.13$ 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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