Problem 39.

A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.2 s later. What is the distance PQ? How far away from the dart board is the dart released?

Solution:

The initial velocity is horizontal, the magnitude is 10 m/s. Then the equation of motion along horizontal axis has the following form:

After 0.2 seconds the dart hits the target. From this condition and from the above equation we can find the distance between the point where the dart is released and the board:

The equations which describe the motion along the vertical axis are the following:

If we substitute into the first equation the traveled time then we can find the distance between points P and Q:

The minus sign means that point Q is below point P.