In this question use $g = 9.8 \, \text{ms}^{-2}$. A ball is projected from a point on horizontal ground with an initial velocity of $7 \text{ms}^{-1}$ at an angle $\theta$ above the horizontal. The ball reaches a maximum vertical height of $h$ metres above the ground. 13 (a) Show that $$h = 2.5 \sin^2 \theta$$ [3 marks] 13 (b) Hence, given that $0^{\circ} \leq \theta \leq 60^{\circ}$, find the maximum value of $h$. [2 marks] 13 (c) Nisha claims that the larger the size of the ball, the greater the maximum vertical height will be. State whether Nisha is correct, giving a reason for your answer. [1 mark]

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In this question use $g = 9.8 \, \text{ms}^{-2}$ - AQA - A-Level Maths Pure - Question 13 - 2022 - Paper 2

Question 13

$In-this-question-use-$g-=-9.8-\,-\text{ms}^{-2}$-AQA-A-Level Maths Pure-Question 13-2022-Paper 2.png$

In this question use $g = 9.8 \, \text{ms}^{-2}$. A ball is projected from a point on horizontal ground with an initial velocity of $7 \text{ms}^{-1}$ at an angle ... show full transcript

Worked Solution & Example Answer:In this question use $g = 9.8 \, \text{ms}^{-2}$ - AQA - A-Level Maths Pure - Question 13 - 2022 - Paper 2

Step 1

13 (a) Show that

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Answer

To find the maximum height $h$ , we can use the kinematic equations. The vertical component of the initial velocity is given by:

$u_y = 7 \sin \theta$ .

The vertical motion can be described using the equation:

$v^2 = u^2 + 2as,$
where:

$v$ is the final vertical velocity (0 m/s at the maximum height),
$u = u_y = 7 \sin \theta$ ,
$a = -g = -9.8 \, \text{ms}^{-2}$ , and
$s = h$ .

Substituting into the formula gives:

$0 = (7 \sin \theta)^2 - 2(9.8)(h).$

Rearranging and solving for $h$ , we find:

$$h = \frac{(7 \sin \theta)^2}{2 \times 9.8} = \frac{49 \sin^2 \theta}{19.6} = 2.5 \sin^2 \theta.$

Step 2

13 (b) Hence, given that 0° ≤ θ ≤ 60°, find the maximum value of h.

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Answer

From the answer in part (a), we have:

$h = 2.5 \sin^2 \theta.$

To maximize $h$ , we need to maximize $\sin^2 \theta$ within the interval $0^{\circ} \leq \theta \leq 60^{\circ}$ . The value of $\sin \theta$ reaches its maximum at $\theta = 60^{\circ}$ :

$\sin 60^{\circ} = \frac{\sqrt{3}}{2}.$

Thus,

$h_{max} = 2.5 \left(\frac{\sqrt{3}}{2}\right)^2 = 2.5 \cdot \frac{3}{4} = 1.875.$

Step 3

13 (c) State whether Nisha is correct, giving a reason for your answer.

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Answer

Nisha is incorrect. The maximum vertical height reached by the ball is determined by its initial velocity and the angle of projection, not the size of the ball. The motion of the ball can be modeled as a particle, and factors such as air resistance and ball size do not influence the vertical height in this ideal scenario.

In this question use $g = 9.8 \, \text{ms}^{-2}$ - AQA - A-Level Maths Pure - Question 13 - 2022 - Paper 2

Worked Solution & Example Answer:In this question use $g = 9.8 \, \text{ms}^{-2}$ - AQA - A-Level Maths Pure - Question 13 - 2022 - Paper 2

13 (a) Show that

13 (b) Hence, given that 0° ≤ θ ≤ 60°, find the maximum value of h.

13 (c) State whether Nisha is correct, giving a reason for your answer.

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