Question

A dart is thrown horizontally with an initial speed of$\mathbf{10}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$toward point$\mathbf{P}$, the bull’s-eye on a dart board. It hits at point$\mathbf{Q}$on the rim, vertically below${}^{\mathbf{P}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{19}\mathbf{s}}$ later. (a)What is the distance${}^{\mathbf{P}\mathbf{Q}}$? (b) How far away from the dart board is the dart released?

Short answer

(a) Distance,${}^{PQ}$is${}^{0.18m}$

(b) Distance of the released dart from the dart board is${}^{1.9m}$

Step-by-step solution

Given information

It is given that,

To understand the concept of free fall

Considering the concept of free fall and the fact that there is no acceleration in the horizontal direction, it is possible to find the distance by which dart moves down, given the time for the horizontal travel.This problem is based on kinematic equations that describe the motion of an object with constant acceleration. Here, the second kinematic equation can be used to find the distances.

Formula:

The displacement in kinematic equation can be written as,

(i)

to find the distance PQ

Vertical distance of the dart is givenby,

(b) to find the distance of the released dart from the dart board

It is given that,

Thus,

$x=1.9m$