﻿
Question
Thu May 05, 2011

# 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?

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
Related Questions
Mon February 20, 2017

# A ball  thrown up is caught by the thrower after 4 seconds . How high did it go and with what velocity was it thrown? How far below its highest point was in 3 second after start? acceleration due to gravity is 9.8.

Mon February 20, 2017

﻿