Question

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

Answer 1

We adopt the positive direction choices used in the textbook so that equations such as
Equation are directly applicable. The initial velocity is horizontal so that $v_{0y}=0$ and $v_{0x}=v_0=10m/s.$
(a) With the origin at the initial point (where the dart leaves the thrower’s hand), the $y$ coordinate of the dart is given by $y=-\frac{1}{2}g{t}^2$, so that $y=-PQ$ we have $PQ=y=\frac{1}{2}\left(9.8m/s^2\right){\left(0.19s\right)}^2=0.18m.$
(b) From $x=v_{0}t$ we obtain $x=\left(10m/s\right)\left(0.19s\right)=1.9m.$