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?

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

where is the vertical distance from P, is the initial velocity, is the acceleration and is time, we have

the answer is b ) square root of 5

double the distance = divide strength by 4