Answered: A ball rolls off a table that is 1.20 m…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

A ball rolls off a table that is 1.20 m high and lands on the floor 1.60 m away from the base of the table. How much time did this take and how fast was the ball going when it rolled off the table?

Expert Solution

Step 1

Given Data:

The height of a table is, $s_{1} = 1.20 m$

The ball rolls from the base of the table at a distance $s_{2}$ is, $s_{2} = 1.60 m$

Let's assume the x-component of the velocity of the ball when it rolled off a table is u_xand the y-component of the velocity is zero $(u_{y} = 0 m/s)$ . The acceleration in the x-direction is zero, and the y-component of the acceleration is equal to the downward acceleration due to gravity. $(a_{y} = - g)$ .

The expression for the second equation of motion is,

$s = u t + \frac{1}{2} a t^{2} . . . . . . . . . . . . . . . . . . . . (1)$

Here, u is the velocity, a is the acceleration and t is the time.

Step 2

From equation (1), evaluate the equation of motion in the x-direction,

$\begin{array}{rcl} s_{2} & = & u_{x} t + \frac{1}{2} a_{x} t^{2} \\ (1.60 m) & = & u_{x} t . . . . . . . . . . . . . . . . . . . . . (2) \end{array}$

Similarly, evaluate the equation of motion in the y-direction,

$\begin{array}{rcl} s_{1} & = & u_{y} t + \frac{1}{2} a_{y} t^{2} \\ (0 m - 1.20 m) & = & \frac{1}{2} \times (- 9.81 {m/s}^{2}) \times t^{2} \\ t^{2} & = & 0.244 s \\ t & = & 0.495 s \end{array}$

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