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Old Jun 19th 2011, 03:08 PM   #1
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Unsure about projectile motion problem

Here's the problem: A basketball is thrown at an angle 55 above the horizontal. It travels 13 ft (along the x axis) before going through the hoop, which is 3 ft above the initial height of the ball.

At what initial speed must the basketball player throw the ball to make the shot?



Now, here is what I've done:
(Note: I originally typed this all in Latex, but after discovering that Latex doesn't work right now, I had to reformat everything. Sorry if it's hard to read)


x = V_{0x}t + 1/2 * at^2

13 = V_{0x}t

t = 13/Vcos(55)


y = V_{0y}t + 1/2 * at^2

3 = Vsin(55) * 13/Vcos(55) - 16 * (13/Vcos(55))^2

3 = 18.57 - 82188/V^2

V = 72.7 ft/s


I'm uncertain of this answer, because I used 13 ft to calculate time relative to the x distance - however, the y distance of 3 feet, I would think, would be referencing the *first* instance the ball hits a height of 3 ft, whereas the x distance occurred on the *second* instance.

Did I make a mistake here, or does it all work out just fine?

Last edited by Sigma; Jun 19th 2011 at 06:21 PM. Reason: Typo
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Old Jun 19th 2011, 06:13 PM   #2
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16 * 13/(Vcos(55))^2

It should be 16 * 13^2/(Vcos(55))^2

Check the above calculation.
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Old Jun 19th 2011, 06:20 PM   #3
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Originally Posted by sa-ri-ga-ma View Post
16 * 13/(Vcos(55))^2

It should be 16 * 13^2/(Vcos(55))^2

Check the above calculation.
That was actually a typo from my stripping out the Latex code. I'll fix it - but as you can see from the following numbers, it was processed correctly.

That aside, how does my approach and result look?
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Old Jun 19th 2011, 08:56 PM   #4
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Originally Posted by Sigma View Post
That was actually a typo from my stripping out the Latex code. I'll fix it - but as you can see from the following numbers, it was processed correctly.

That aside, how does my approach and result look?
You have missed a decimal.
16*13^2/(cos55)^2 = 8219.09
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Old Jun 19th 2011, 09:07 PM   #5
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Originally Posted by sa-ri-ga-ma View Post
You have missed a decimal.
16*13^2/(cos55)^2 = 8219.09
Thank you for spotting that!

That said, is there an issue conceptually with this, given what I mentioned in my initial post?
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Old Jun 20th 2011, 04:47 AM   #6
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Originally Posted by Sigma View Post
Thank you for spotting that!

That said, is there an issue conceptually with this, given what I mentioned in my initial post?
In the first instance the ball is going up. So it can't go through the hoop.
In the second instance it is coming down. So it can go through the hoop.
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Old Jun 20th 2011, 05:04 AM   #7
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Originally Posted by sa-ri-ga-ma View Post
In the first instance the ball is going up. So it can't go through the hoop.
In the second instance it is coming down. So it can go through the hoop.

Oh, there's no question that the second instance is what I want to reference. The issue for me is that the equation I set up simply references "3" and there are two cases where the y displacement is 3. My uncertainty is whether the equation will reference the proper one, or if I'm getting an erroneous answer because I needed to do things differently.

Are you saying that my setup is correct?
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Old Jun 20th 2011, 09:19 AM   #8
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Don't worry, it is correct. If you're not sure, just use the equations:

h = ut + 0.5at^2

and

d = ut

to get the time in both cases. If they are the same the answer is good. But concerning your question, it was either this or the negative of the velocity, so don't worry
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Old Jun 21st 2011, 07:43 PM   #9
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Originally Posted by Unknown008 View Post
Don't worry, it is correct. If you're not sure, just use the equations:
Thanks! I appreciate the help.
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