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
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 $75 \frac{\pi}{180} = \frac{5}{12}\pi$
$$\sqrt{\frac{78400}{\sin\left(\frac{5}{12}\pi\right)\cos\left(\frac{5}{12}\pi\right)}} = \dots $$