# A dart is thrown horizontally with an initial speed of 18 m/s toward point P, the bull's-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.19 s later. (a) What is the distance PQ? (b) How far away from the dart board ?

Oct 31, 2015

$\left(a\right) . 17.7 \text{cm}$

$\left(b\right) . 3.42 \text{m}$

#### Explanation:

(a). To find PQ we consider only the vertical component of the motion.

Since $u$ is zero we can write:

$s = \frac{1}{2} \text{g} {t}^{2}$

$s = \frac{1}{2} \times 9.8 \times {0.19}^{2}$

$s = 0.177 \text{m"=17.7"cm}$

(b).

The horizontal component of velocity is constant so:

Distance $= v \times t$

$= 18 \times 0.19 = 3.42 \text{m}$