Question

A dart is thrown horizontally toward the bull's-eye, point P on the dart board in the figure, with an initial speed of 9.6 m/s. It hits at point Q on the rim, vertically below P 0.17 s later. What is the distance PQ? How far away from the dart board did the dart thrower stand?

Answer 1

What is the distance PQ? (vertical motion)

You do not need any angle to mess with in this problem.

Let

Y = distance PQ and this is determined by the working formula

Y = Vy(T) + (1/2)gT^2

where

Vy = vertical component of the initial velocity = 0
T = time of flight = 0.17 sec.
g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Substituting values,

Y = 0 + (1/2)(9.8)(0.17)^2

Y = 0.141 m = 14.1 cm.


How far away from the dart board did the dart thrower stand? (horizontal motion)

Working formula is

X = Vx(T)

where

X = distance of thrower from the dart board
Vx = horizontal component of the initial velocity = 9.8 (given)
T = 0.17sec

and substituting appropriate values,

X = 9.8(0.17)

X = 1.66 meters = 166 cm.