Question

A dart is thrown horizontally with an initial speed of 10m/s toward point P,the bull's eye on a dart board.It hits at point Q on the rim,vertically below P,0.19s later. What is the distance PQ? How far away from the dart board is the dart released?

Answer 1

0.177 meters between P and Q.
The dart was released 1.9 meters from the board.

Explanation:

Since we know that the dart was initially traveling horizontally and we know that the time of flight was `0.19` seconds, the only force acting in the vertical direction is gravity. So we need to calculate the distance an object would fall in this time.

The equation for this is:
`d = 1/2 at^2`
where `d` is the vertical distance traveled at time `t` and `a` is the acceleration due to gravity.

`d = 1/2*9.8*0.19^2`
`d = 0.177` meters

The distance between the initial release point and the dart board may be calculated more easily. Since gravity does not act in the horizontal direction, the speed will remain constant. Since we know the initial speed `10 m/s` and the time of flight `0.19 s` we need only to solve this simple equation:
`D = vt`
`D = 10 m/s * 0.19 s = 1.9 m`