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Diagram of the question: https://i.stack.imgur.com/f0NfR.jpg

The question I have is part (ii) and (iii)

A particle is projected from a point $O$ with speed $Vms^{−1}$ at an acute angle $\alpha$ above the horizontal. Acceleration due to gravity is $10ms^{−2}$ down. At time $t$ seconds it has horizontal and vertical displacements $x$ metres and $y$ metres respectively from $O$. At point $P$ where it attains its greatest height, the angle of elevation from $O$ is $β$.

(ii) Show that $\tan β=\frac{1}{2}\tan \alpha$

(iii) If the particle has greatest height $80m$ above $O$ at a horizontal distance $120m$ from $O$, find the exact value of $\alpha$ and $V$.

bjcolby15
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Just_A_User
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1 Answers1

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ii) The time to reach the highest point is $t=\frac{v_y}g$. The horizontal and the vertical distances travelled are, respectively,

$$x=v_xt=\frac{v_xv_y}g,\>\>\>\>\>y=v_yt -\frac12 gt^2= \frac{v_y^2}{2g}$$

As a result,

$$\tan\beta = \frac yx= \frac12 \frac{v_y}{v_x}=\frac12\tan\alpha$$

iii) Given $x=120$ and $y=80$, we get $\tan\beta = \frac{80}{120}=\frac23$ and

$$\alpha = \tan^{-1} (2\tan\beta)= \tan^{-1} \frac43,\>\>\>\>\>\sin\alpha =\frac45$$

$$V= \frac{v_y}{\sin\alpha}= \frac{\sqrt{2gy}}{\sin\alpha}=50 m/s$$

Quanto
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