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MaZnFLiP

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## Homework Statement

A box of mass 4.750Kg is suspended from a rope which is attached to the ceiling and is currently at rest. You then shoot an arrow horizontally with a velocity of 21.00 m/s toward the box while standing 4.5m away from it. You may neglect air resistance.

A. With what velocity will the arow strike the box?

B. If the arrow has a mass of 0.2500Kg, what is the max change in height the box theoretically will experience?

## Homework Equations

b = Box, a = arrow

KEib + KEfb = KEia + KEfa

Fc = MV^2/r

Fg = mg

## The Attempt at a Solution

So what I did was I first solved for the Final Velocity of the box which would become the new initial velocity after the arrow struck the box using the KEib + KEfb = KEia + KEfa.

0.5(4.750)V^2 = 0.5(0.25)(21.00)^2

Solving for the Velocity, I got 4.82m/s. so that becomes the new velocity I have and the New initial velocity would be 0. My New givens are:

Vi = 4.82m/s

Vf = 0m/s

A = -9.81m/s^2

[tex]\Delta[/tex]y = ?

If I'm looking for the change in height, I don't feel like I need Centripetal Force. Would I just need to worry about [tex]\Delta[/tex]y? or do I have to use the Centripetal force equation?

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