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A particle $P$ is projected from a point $O$ with initial speed $10{\text{ }}m{\text{ }}{s^{ - 1}}$ at an angle of ${45^ \circ }$ above the horizontal. $P$ subsequently passes through the point $A$ which is at an angle of elevation of ${30^ \circ }$ from $O$ (see diagram). At time $t\,s$ after projection the horizontal and vertically upward displacements of $P$ from $O$ are $x$ m and $y{\text{ }}m$ respectively.

a) Write down expressions for $x$ and $y$ in terms of $t$, and hence obtain the equation of the trajectory of $P$.

b) Calculate the value of $x$ when $P$ is at $A$.

c) Find the angle the trajectory makes with the horizontal when $P$ is at $A$.

پاسخ تشریحی :
نمایش پاسخ

a) $x = \left( {10\cos {{45}^ \circ }} \right)t$ and

$y = \left( {10\sin {{45}^ \circ }} \right)t - g{t^2}/2$

$y = \left( {10\sin {{45}^ \circ }/{\text{ }}10\cos {{45}^ \circ }} \right)x - 10{\left( {x/10\cos {{45}^ \circ }} \right)^2}/2$

$y = x - {x^2}/10$

b) $y/x = \tan {30^ \circ }$

$1 - x/10 = \tan {30^ \circ }$

$x = 4.23$

c) $dy/dx = 1 - 2x/10$

$\tan \theta  = dy/dx$

$\tan \theta  = 1 - 2 \times 4.23/10\left( { = 0.15472..} \right)$

$\theta  = {8.79^ \circ }$

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