# Trouble with caculus problem, parametric equations, I don't know what I'm doing wrong

When a mortar shell is ﬁred with an initial velocity of v0 ft/sec at an angle α above the horizontal, then its position after t seconds is given by the parametric equations $x = (v0 \cos \alpha)t$ , $y = (v0 \sin \alpha)t − 16t^2$

If the mortar shell hits the ground 4900 feet from the mortar when α = 75◦, determine v0.

So I've tried various forms of: \begin{align*} t = {} & 4900/(v0 \cos 75) \\ 0 = {} & (v0 \sin 75)(4900/(v0 \cos 75)) - 16(4900/(v0 \cos 75))^2 \\ 4900(v0 \sin 75)/(v0 \cos 75) = {} & 384160000/(v0 \cos 75)^2 \\ v0 \sin 75 = {} & 78400/(v0 \cos 75) \\ v0 = {} & 78400/\sin 75 * v0 * \cos 75 \\ v0^2 = {} & 78400/\sin 75 * \cos 75 \\ v0 = {} & 468.33...i \end{align*}

which doesn't seem right. And the answer choices are:

1. v0 = 530 ft/sec
2. v0 = 560 ft/sec
3. v0 = 520 ft/sec
4. v0 = 550 ft/sec
5. v0 = 540 ft/sec
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$$\sqrt{\frac{78400}{\sin75\cos75}} = 468.3359976i$$

However, if we convert $75^\circ$ to radians $75 \frac{\pi}{180} = \frac{5}{12}\pi$

$$\sqrt{\frac{78400}{\sin\left(\frac{5}{12}\pi\right)\cos\left(\frac{5}{12}\pi\right)}} = \dots$$

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Oh crap! Yep, it was in radians. Thank you :] – clinton Oct 10 '13 at 18:32

Using the given information we get: $$4900=v_0*t*cos75$$ $$16t=v_0sin75$$ Multiplying these equations we get: $$4900*16*t={v_0}^2*cos75*sin75*t$$ So $$4900*16={v_0}^2*{1\over4}$$ Hence $$v_0=560$$

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Set $y=0$ and solve for $t$. The solutions are $t=0$ (when the shell was fired) and $t=\frac{v_0\sin\alpha}{16}$.

Substitute this value of $t$ in the expression for $x$. We get $$x=\frac{v_0^2\sin\alpha\cos\alpha}{16}.$$ Set $x$ equal to $4900$, and solve for $v_0$.

Remarks: We don't even need a calculator. For $\sin\alpha\cos\alpha=\frac{1}{2}\sin 2\alpha=\frac{1}{4}$.

Traditionally, artillery officers had some training in mathematics. This was particularly the case in France. Napoleon knew some mathematics. There is even a theorem in geometry that is sometimes credited to him.

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