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Old Sep 14th 2010, 10:43 PM   #1
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"Win the Prize" Projectile motion

Hello all,
I have worked through this problem several different ways, and cannot get my answer to match that of the book's. The attached picture is my work thus far, if you could, please point out any mistakes or hints as to where I must go with my arithmetic.

Thank you,
CC
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Old Sep 14th 2010, 10:48 PM   #2
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Also, please ignore the random 1.29m/s in the middle of the graph. I forgot to post the question as well. Here it is:

In a carnival booth, you win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a shelf about the point where the quarter leaves your hand and is a horizontal distance of 2.1m from this point. If you toss the coin with a velocity of 6.4m/s at an angle of 60(degrees) above the horizontal, the coin lands in the dish. You can ignore air resistance. (a) what is the height of the shelf above the point where the quarter leaves your hand? (b) What is the vertical component of the velocity of the quarter just before it lands in the dish?
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Old Sep 14th 2010, 10:56 PM   #3
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Here is the picture that the book provides.
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Old Sep 15th 2010, 07:16 AM   #4
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Here's how to proceed:

1. The time of the flight is D_x/v_x, where D_x = horizontal distance traveled = 2.1m, and v_x = horiz velocity = 6.4 m/s cos(60) = 3.2 m/s. Hence t = 2.1m/(3.2m/s) = 0.66s

2. The vertical position of the coin at t=0.66s is:

y = ut + 1/2 at^2 = 6.4m/s sin(60) 0.66s - 1/2(9.8 m/s^2)(0.66s)^2 = 3.64m

The vertical velocity at t= 0.66s is:

v = u + at = 6.4m/s sin(60) - 9.8m/s^2 (0.66s) = -0.89m/s
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Old Sep 15th 2010, 08:30 AM   #5
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Thank you very much, I found out where I goofed! It is much appreciated
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Old Sep 15th 2010, 08:36 AM   #6
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So - what prize do I win????
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Old Sep 15th 2010, 08:53 AM   #7
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A thanks!
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Old Sep 15th 2010, 04:02 PM   #8
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Originally Posted by ChipB View Post

v = u + at = 6.4m/s sin(60) - 9.8m/s^2 (0.66s) = -0.89m/s
Hello again, I was trying to find the Vy at .66s like you do shown in the quote above. When I do the calculation myself I get around -.93m/s

Also, will you explain what you did there?

What I see is this:

Vy-at

Where did you derive that equation from?

Sorry for all of the questions, I am just trying to see where I am missing something.

Thanks,
CC
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Old Sep 15th 2010, 06:00 PM   #9
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I think the difference between your answer and mine is I used t = 2.1m/(6.4 m/s x cos(60)) = 2.1/3.2 = 0.65625s , and you may have rounded this to 0.66s.

v(t= 0.656s) = 6.4 sin(60) - (9.8)(0.656) = -0.89m/s

This equation comes from:

v = u +at

where u = initial velocity and a = acceleration. Here a = -g = -09.8 m/s^2.
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Old Dec 30th 2017, 01:48 AM   #10
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Originally Posted by CalculusCrazed View Post
Here is the picture that the book provides.
please may i know the name of the book
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