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Old Oct 18th 2009, 11:12 PM   #1
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2 Questions from Sample Midterm, Help!

Hey guys,
midterm's coming up and the sample has some pretty intense questions that I'm not really getting. Just wondering if you could help me out with process..

1.A rain drop falls vertically down beside a tall building. It takes 0.5s for the rain drop to pass by the second floor window, which is 1.5 m tall. If the bottom of the window is 3.5m from the ground, find the time for the raindrop to go from the bottom of the window to the ground. Neglect air resistance.

answer given: 0.456 s

2.A basketball player makes a shot from a height of 2.50 m at an angle of 30 degrees above the horizontal. The horizontal distance between the player and the hoop is 6.00 m. The hoop is 3.00m above the ground. Find the magnitude of the initial velocity of the ball which will cause it to go into the basket is.

answer given: 8.91 m/s

Anything you could tell me about the process would be appreciated..
Thanks!

Steve
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Old Oct 19th 2009, 12:39 AM   #2
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Welcome to the forum. You have to attempt to solve the question and post your attempt before we can help. Or if you have any conceptual difficulty, you should specify that.
In the meanwhile, for starters, you can check the previous threads in this section. Many similar problems have been solved.
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Old Oct 19th 2009, 08:52 AM   #3
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Talking

Oh, sorry. I didn't realize we had to post attempts.
Alright.

For the raindrop question I tried to use a series of the 4 kinematic equations:

v=vi+at
v^2=vi^2+2ad
d=(vi)t+(1/2)at^2
d=[(vi+v)/2]t

I lost the sheet that I tried to do it on, but I used d=(vi)t+(1/2)at^2 (d=1.5 a=-9.81 t=.5s) to find the initial velocity at the top of the window. Then inserted that value into v^2=vi^2+2ad (a=-9.81 d=1.5) to find the velocity at the bottom of the window. I then took that value (velocity at bottom of window frame) inserted it into v^2=vi^2+2ad (d=3.5 a=-9.81) to solve for the velocity right before ground impact. Then used that speed to calculate the time taken to travel the 3.5 meters.
Am I on the right track?


for the basketball player one (the real challenge) I didn't even have a clue where to begin.

All you have is an arbitrary angle and another co-ordinate on the parabolic arc that the ball follows. But I'm sure you're not supposed to use any quad-formula (I'm in first year Uni physics and we haven't even mentioned quad-form.. also I can't remember it..) You're given no value for time, speed, maximum height, x or y component velocities. Any work I've done has just been experimenting with different equations, nothing has worked so far.

please help me!
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Old Oct 19th 2009, 11:06 AM   #4
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Ok, about the above. I tried to work through the raindrop question again.
Here's what happened. Please critique:


1.5m=vi(0.5)+1/2(9.81)(0.25)
vi=.5476

vf^2=(.5476)^2+2(9.81)(1.5)
vf=1.747045 (velocity at bottom of window)

vf^2=(1.747045)^2+2(9.81)(3.5)
vf=8.4689

3.5=[(.747045+8.7689)/2]t
t=.66 rpt

It's the wrong answer! please help.
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Old Oct 19th 2009, 11:36 AM   #5
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To find v at bottom of 1.5 m window
Taking (a) = 9.8 m/s^2
= 1.5/t + (1/4a) = 3 + 2.45 = 5.45 m/s

v becomes u at bottom of window

using s = ut + (1/2)at^2 = 5.45 x 0.456 + (4.9 x 0.456^2) = 3.5 m

Total s = 5.015 m
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Old Oct 19th 2009, 12:58 PM   #6
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Sorry, I don't understand what you calculated..
What is s?

And what is the original equation you used to calculate the velocity at the bottom of the window?
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Old Oct 19th 2009, 02:28 PM   #7
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Originally Posted by Cutcutler View Post
Sorry, I don't understand what you calculated..
What is s?

And what is the original equation you used to calculate the velocity at the bottom of the window?
s = displacement

distance = speed x time
speed therefore = distance/ time = 1.5m/0.5 secs = 3 m/s

since the drop is accelerating, the 3 m/s will be the average speed over the distance, so 3 m/s will be the speed at the middle of the window.

To find the velocity at the bottom of window you use the time/2 x acceleration (a) + average velocity = (0.5/2 x 9.8) + 3 = 5.45 m/s

v = u + at, since u is zero because the drop fell, you can find time by v/a = t.

s = ut + (1/2)at^2 = 0 + (4.9 x 0.5561^2) = 1.5154 m from where drop fell to bottom of window.
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Old Oct 19th 2009, 06:29 PM   #8
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We don't know what the velocity at the top of the window is.

The questions says that the drop fell from the top of the building, and it passed by the second story window in .5 seconds. It doesn't say anything about starting from rest.
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Old Oct 20th 2009, 09:13 AM   #9
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Oh yes i do know what the velocity at top of window is!!

It will be 3 m/s -(t/2 x 9.8) = 0.55 m/s

I'll keep a note of your name, because i won't be helping you out again!

also if you expect help from members on here i suggest you learn how to click the thanks button.

p.s also read your question properly:

It takes 0.5s for the rain drop to pass by the second floor window, which is 1.5 m tall

Last edited by Paul46; Oct 20th 2009 at 09:16 AM.
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Old Oct 20th 2009, 06:00 PM   #10
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Excuse me, I wasn't aware that questioning an answer was in some way offensive.

If I am questioning information you present me with it is because I don't understand it, not because I hold a personal vendetta against you or any other kind people who help students understand the concepts that they're just learning about.

In this case, the line in your previous response:

"v = u + at, since u is zero because the drop fell, you can find time by v/a = t"

suggested to me that you thought the velocity at the top of the window was 0, when in fact this isn't true. Now, I may be totally incorrect in saying that, but I'd rather you handled it in a mature manner and told me what it was that you had actually meant rather than assuming that I'm out to get you.

I feel that the way you responded was very rude and childish in nature.
A human being is not something to be dictated to, I learn by asking questions and if you're not ready to be questioned, don't teach.

Thanks button,
Steve
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