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:


*

*v0 = 530 ft/sec

*v0 = 560 ft/sec

*v0 = 520 ft/sec

*v0 = 550 ft/sec

*v0 = 540 ft/sec

 A: 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.
A: You seem to have your calculator set to radians, but are entering your angles in degrees.
$$\sqrt{\frac{78400}{\sin75\cos75}} = 468.3359976i $$
However, if we convert $75^\circ$ to radians $75 \frac{\pi}{180} = \frac{5}{12}\pi$
$$\sqrt{\frac{78400}{\sin\left(\frac{5}{12}\pi\right)\cos\left(\frac{5}{12}\pi\right)}} = \dots $$
A: 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 
$$v_0=560$$
