Projectile Motion (1 Viewer)

michaeljennings · Jun 14, 2015

Does anyone have the solution to Chapter 3H Question 11b from the Cambridge Year 12 Textbook? Cheers

InteGrand · Jun 14, 2015

michaeljennings said:
Does anyone have the solution to Chapter 3H Question 11b from the Cambridge Year 12 Textbook? Cheers

$Cartesian equation of flight path is $x^2\tan^2 \alpha -(4kx)\tan \alpha + (4ky + x^2) = 0$, where $k\equiv \frac{V^2}{2g}$ and $\alpha$ is acute (part (a) information).$

$The projectile passes through $(X,Y)$ in the first quadrant $\Longleftrightarrow X^2\tan^2 \alpha -(4kX)\tan \alpha + (4kY + X^2) = 0$. This is a quadratic equation in $\tan \alpha$, so there will be two distinct real solutions for $\tan \alpha$ (and hence two different angles, since $\tan$ is one-to-one for acute arguments) only if the discriminant of the quadratic is positive, i.e. $\Delta = (4kX)^2 - 4(X^2)(4kY + X^2)> 0 \Longleftrightarrow 16k^2 X^2 - 16kX^2 Y - 4(X^2)^2>0$$

$$\Longleftrightarrow 4k^2-4kY - X^2 >0$ (dividing through by $4X^2$)$

$$\Longleftrightarrow X^2 < 4k^2 -4kY$, as required.$

michaeljennings · Jun 14, 2015

InteGrand said:
$Cartesian equation of flight path is $x^2\tan^2 \alpha -(4kx)\tan \alpha + (4ky + x^2) = 0$, where $k\equiv \frac{V^2}{2g}$ and $\alpha$ is acute (part (a) information).$

$The projectile passes through $(X,Y)$ in the first quadrant $\Longleftrightarrow X^2\tan^2 \alpha -(4kX)\tan \alpha + (4kY + X^2) = 0$. This is a quadratic equation in $\tan \alpha$, so there will be two distinct real solutions for $\tan \alpha$ (and hence two different angles, since $\tan$ is one-to-one for acute arguments) only if the discriminant of the quadratic is positive, i.e. $\Delta = (4kX)^2 - 4(X^2)(4kY + X^2)> 0 \Longleftrightarrow 16k^2 X^2 - 16kX^2 Y - 4(X^2)^2>0$$

$$\Longleftrightarrow 4k^2-4kY - X^2 >0$ (dividing through by $4X^2$)$

$$\Longleftrightarrow X^2 < 4k^2 -4kY$, as required.$

Ah I overlooked that, cheers!

Projectile Motion (1 Viewer)

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InteGrand

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