I need assistance solving this question 4. Bob tosses a pen straight up into the air, but does not catch it, so it falls to the floor. The speed of thepen is 3.7 m/s when it leaves his hands, and his hands are 1.2 m above the floor.a) What is the sign of the pen's velocity on the way up?What about the sign of its acceleration on the way up?b) How long was the pen in the air?c) What is the pen's velocity just before it hits the floor?...at the top?...at the top?...on the way down?...on the way down?

Name: I need assistance solving this question 4. Bob tosses a pen straight u
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Description: I need assistance solving this question 4. Bob tosses a pen straight up into the air, but does not catch it, so it falls to the floor. The speed of thepen is 3.7 m/s when it leaves his hands, and his hands are 1.2 m above the floor.a) What is the sig

Answered: 4. Bob tosses a pen straight up into…

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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I need assistance solving this question

4. Bob tosses a pen straight up into the air, but does not catch it, so it falls to the floor. The speed of the
pen is 3.7 m/s when it leaves his hands, and his hands are 1.2 m above the floor.
a) What is the sign of the pen's velocity on the way up?
What about the sign of its acceleration on the way up?
b) How long was the pen in the air?
c) What is the pen's velocity just before it hits the floor?
...at the top?
...at the top?
...on the way down?
...on the way down?

Expert Solution

Step 1

a) Sign of the pen's velocity on the way up = Positive

Sign of the pen's velocity on the top = Positive

Sign of the pen's velocity on the way down = Negative

Acceleration due to gravity always works downward, so direction of accelration is downward, hence the sign is always negative

Sign of the pen's acceleration on the way up/on the top/on the way down is Negative.

During upward journey,

Initial velocity, u = 3.7 m/s

Velocity at the top, v = 0

Acceleration, a=g=-9.8 m/s²

Time,

$\begin{array}{rcl} t_{1} & = & \frac{v - u}{a} \\ = & \frac{0 - 3.7}{- 9.8} \\ = & 0.378 s \end{array}$

Distance covered,

$\begin{array}{rcl} v^{2} - u^{2} & = & 2 g s_{1} \\ \Rightarrow & s_{1} = \frac{v^{2} - u^{2}}{2 g} \\ = & \frac{0 - 3 . 7^{2}}{- 2 \times 9.8} \\ = & 0.698 m \end{array}$

During downward journey,

Initial velocity, u = 0 m/s

Velocity just before it hits ground, v = ?

Distance covered, S = 0.698+1.2 =1.898 m

Acceleration, a=g=9.8 m/s²

^Hence,

$\begin{array}{rcl} v^{2} - u^{2} & = & 2 g S \\ \Rightarrow & v^{2} - 0 = 2 g S \\ \Rightarrow & v = \sqrt{2 g S} \\ \Rightarrow & v = \sqrt{2 (9.8) 1.898} \\ \Rightarrow & v = 6.099 m / s \end{array}$

$\begin{array}{rcl} t_{2} & = & \frac{v - u}{g} \\ = & \frac{6.099 - 0}{9.8} \\ = & 0.622 s \end{array}$

SO,

The pen was in air for (0.378+0.622)=1 second

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