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

0.18 m

Step-by-step 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+(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*1.9+(1)/(2)9.8*0.19^2 = 0.17689 = 0.18 \text{ m}`