# An arrow is shot at a 30 degree angle with a velocity of 39m/s. How high will it go? What horizontal distance will it travel?

Dec 4, 2015

(a)

$19.4 \text{m}$

(b)

$134.4 \text{m}$

#### Explanation:

(a)

Considering the vertical component ot the motion:

${v}^{2} = {u}^{2} + 2 a s$

This becomes:

${v}^{2} = {u}^{2} - 2 g h$

$\therefore 0 = {\left(39 \sin 30\right)}^{2} - 2 \times 9.8 \times h$

$\therefore 0 = 380 - 19.6 \times h$

$h = \frac{280}{19.6} = 19.4 \text{m}$

(b)

The expression for range $d$ on level ground is:

$d = \frac{{v}^{2} \sin \left(2 \theta\right)}{g}$

$\therefore d = \frac{{39}^{2} \times \sin \left(60\right)}{9.8}$

$d = 134.4 \text{m}$