Question
Thu May 05, 2011 By: Aurghyadip Kundu

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

Expert Reply
Fri May 06, 2011
x = x0 + v0x t ? x = 10 m/st vx = v0x ? vx = 10 m/s

y = y0 + v0y t ? 1/2 gt2 ? y = ? 1/2 gt2  
 
vy = v0y ? gt ? vy = ?gt
 
(a) Because we know the time the dart is in ?ight, we can ?nd the y value corre-sponding to this time:

y = ? 1/2 gt2 = ?1/2 (9.8 m/s2 )(0.19 s)2
 
y = ?0.177 m = ?17.7 cm

The dart hits 17.7 cm below the bull’s eye.
 
b) Knowing the time of ?ight, we can also ?nd the total distance to the board:
 
x = 10 m/st = 10 m/s(0.19 s) = 1.9 m
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