A dart is thrown horizontally with an initial speed of 10 m/s towards point P, the bull's eye on a dart board. It hits at the point Q on the rim, vertically below P, 0.2 seconds later, what is distance PQ?
0.2 m
Consider Vertical motion of dart
uy=0ay=−gs=ut+12 at2t=0.2 secs=0−12×10×0.2×0.2s=−0.2 m.