# Projectile motion with trig? (Gen Physics 1 question)

## An archer shoots an arrow at a target that is a horizontal distance d = 55 m away; the bull’s-eye of the target is at same height as the release height of the arrow. At what angle, in degrees above the horizontal, must the arrow be released to hit the bull’s-eye if the arrow's initial speed is 39 m/s?

Feb 22, 2017

A useful expression to use for the range is:

$\textsf{d = \frac{{v}^{2} \sin 2 \theta}{g}}$

$\therefore$$\textsf{\sin 2 \theta = \frac{\mathrm{dg}}{{v}^{2}}}$

$\textsf{\sin 2 \theta = \frac{55 \times 9.81}{39} ^ 2}$

$\textsf{\sin 2 \theta = 0.3547}$

$\textsf{2 \theta = {20.77}^{\circ}}$

$\textsf{\theta = {10.4}^{\circ}}$

Feb 22, 2017

${15.65}^{\circ}$

#### Explanation:

The parabolic path described by the arrow considering the coordinates origin at the archer position, is

$\left(x , y\right) = \left({v}_{0} \cos \theta t , {v}_{0} \sin \theta t - \frac{1}{2} g {t}^{2}\right)$

After ${t}_{0}$ seconds the target is hit so

${v}_{0} \cos \theta {t}_{0} = d \to {t}_{0} = \frac{d}{{v}_{0} \cos \theta}$

at this time ${t}_{0}$ also

${v}_{0} \sin \theta {t}_{0} - \frac{1}{2} g {t}_{0}^{2} = 0$ or substituting

${v}_{0} \sin \theta \left(\frac{d}{{v}_{0} \cos \theta}\right) - \frac{1}{2} g {\left(\frac{d}{{v}_{0} \cos \theta}\right)}^{2} = 0$

Simplifying

${v}_{0}^{2} \sin \theta \cos \theta - \frac{1}{2} g {d}^{2} = 0$ or

$2 \sin \theta \cos \theta = \sin \left(2 \theta\right) = \frac{g d}{v} _ {0}^{2}$ and finally

$\theta = \frac{1}{2} \arcsin \left(\frac{g d}{v} _ {0}^{2}\right) = \frac{1}{2} \arcsin \left(9.81 \left(\frac{55}{39} ^ 2\right)\right) = 0.273148$[rad] = ${15.65}^{\circ}$