Question

a dart is thrown horizontally with an initial speed of 16 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 is the dart released? (a) number \boxed{} units (b) number \boxed{} units

Explanation:

Step1: Vertical distance (PQ) calculation

We use the free-fall displacement formula for vertical motion (initial vertical velocity $v_{0y}=0$):
$$\Delta y = v_{0y}t + \frac{1}{2}gt^2$$
Substitute $v_{0y}=0$, $g=9.8\ \text{m/s}^2$, $t=0.19\ \text{s}$:
$$PQ = 0 + \frac{1}{2} \times 9.8 \times (0.19)^2$$
$$PQ = 4.9 \times 0.0361 = 0.17689\ \text{m}$$

Step2: Horizontal distance calculation

Horizontal motion has constant velocity, so use $d = v_{0x}t$:
Substitute $v_{0x}=16\ \text{m/s}$, $t=0.19\ \text{s}$:
$$d = 16 \times 0.19 = 3.04\ \text{m}$$

Answer:

(a) Number: 0.18 (rounded to 2 significant figures) Units: meters
(b) Number: 3.04 Units: meters