A ball is projected forward from a fixed point, P, on a horizontal surface with an initial speed $u \text{ ms}^{-1}$, at an acute angle $\theta$ above the horizontal. The ball needs to first land at a point at least $d$ metres away from P. You may assume the ball may be modelled as a particle and that air resistance may be ignored. Show that $$\sin 2\theta \geq \frac{dg}{u^2}$$

A-Level AQA Maths Pure Modelling with Functions: A ball is projected forward from

A ball is projected forward from a fixed point, P, on a horizontal surface with an initial speed $u \text{ ms}^{-1}$, at an acute angle $\theta$ above the horizontal - AQA - A-Level Maths Pure - Question 17 - 2020 - Paper 2

Question 17

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A ball is projected forward from a fixed point, P, on a horizontal surface with an initial speed $u \text{ ms}^{-1}$, at an acute angle $\theta$ above the horizontal... show full transcript

Worked Solution & Example Answer:A ball is projected forward from a fixed point, P, on a horizontal surface with an initial speed $u \text{ ms}^{-1}$, at an acute angle $\theta$ above the horizontal - AQA - A-Level Maths Pure - Question 17 - 2020 - Paper 2

Step 1

Model vertical motion using a suitable constant acceleration equation

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Answer

We begin by considering the vertical motion of the ball. The vertical displacement $y$ can be expressed in terms of time $t$ as follows:

$y = ut \sin \theta - \frac{1}{2}gt^2$

At the maximum height, we set $y = 0$ . Thus:

$0 = ut \sin \theta - \frac{1}{2}gt^2$

This can be rearranged to find time $t$ :

$t = \frac{2u \sin \theta}{g}$

Step 2

Model horizontal displacement

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Answer

Next, we analyze the horizontal motion of the ball. The horizontal displacement $x$ is given by:

$x = u t \cos \theta$

Substituting for $t$ from our previous step:

$x = u \left( \frac{2u \sin \theta}{g} \right) \cos \theta$

Thus:

$x = \frac{2u^2 \sin \theta \cos \theta}{g}$

Step 3

Show that horizontal distance $x$ is at least $d$

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Answer

For the ball to land at least $d$ metres away, we set the inequality:

$\frac{2u^2 \sin \theta \cos \theta}{g} \geq d$

Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$ , we can rewrite this as:

$\frac{u^2 \sin 2\theta}{g} \geq d$

This simplifies to:

$\sin 2\theta \geq \frac{dg}{u^2}$

A ball is projected forward from a fixed point, P, on a horizontal surface with an initial speed $u \text{ ms}^{-1}$, at an acute angle $\theta$ above the horizontal - AQA - A-Level Maths Pure - Question 17 - 2020 - Paper 2

Worked Solution & Example Answer:A ball is projected forward from a fixed point, P, on a horizontal surface with an initial speed $u \text{ ms}^{-1}$, at an acute angle $\theta$ above the horizontal - AQA - A-Level Maths Pure - Question 17 - 2020 - Paper 2

Model vertical motion using a suitable constant acceleration equation

Model horizontal displacement

Show that horizontal distance $x$ is at least $d$

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