Question

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, 0.19 s later. (a) What is the distance ? (b) How far away from the dart board is the dart released?

Answer 1

then PQ = \[\frac{ 1 }{ 2 }g t ^{2}\]
=\[1/2 * 9.8*(0.19^{2})\]
=0.1769 m = 17.69 cm

Answer 2

distance of the dart board = 0.19* 10 = 1.9 m

Answer 3

The distance PQ is the distance the dart DROPS during it's flight of 0.19 s. Let D=magnitude of PQ.
D = (1/2)g t^2 where g = 9.8m/sec^2 and t = .19s =
= (1.2)(9.8) (0.19^2) = 0.94 m

Assuming no air resistance, the distance = (19. m/s) (.19 s) = 3.61 m