Question

A dart is thrown horizontally with an initial speed of 16 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. What is the distance P Q?

Answer 1

Answer:

0.18 m

Explanation:

The motion of the dart comprises horizontal and vertical components. Vertically, the initial velocity is 0 because the dart was thrown horizontally. Also, the vertical motion is under gravity. Using the equation of motion

$s = ut+\frac{1}{2}at^2$

where $s$ is the vertical distance from P, $u$ is the initial velocity, $a$ is the acceleration and $t$ is time, we have

$s = 0\times1.9+\frac{1}{2}9.8\times0.19^2 = 0.17689 = 0.18 \text{ m}$