A dart is thrown horizontally with an initial speed of 18 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 PQ? (b) How far away from the dart board ?

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Answer 1 · 2015-10-31T18:03:53

#(a). 17.7"cm"#

#(b). 3.42"m"#

(a). To find PQ we consider only the vertical component of the motion.

Since #u# is zero we can write:

#s=1/2"g"t^2#

#s=1/2xx9.8xx0.19^2#

#s=0.177"m"=17.7"cm"#

(b).

The horizontal component of velocity is constant so:

Distance #= vxx t#

#=18xx0.19=3.42"m"#