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9/22/2017 12:11:25 AM
Posted: 10/5/2005 8:46:03 PM EDT
[Last Edit: 10/5/2005 8:47:44 PM EDT by Sparsky]
Need help with a problem. I'm just tring to understand how to get the answer, etc.

If I were to kick a football at 10m/sec at a 35 degree angle at an object 4 meters away and 1.5 meters high would I hit the object or over shoot it? How high or low would I be and how far would the ball travel if I miss the object? Force of gravity being 9.80m/sec squared.

If anyone knows this in "dummy" terms please LMK ASAP. Thanks

Edit to add: Weight of object unknown/not in factor along with air resistance.
Link Posted: 10/5/2005 8:50:00 PM EDT
[Last Edit: 10/5/2005 8:56:42 PM EDT by ProfessorEvil]
10m/s * cos (35) * time in seconds = distance at time x

How long does it take to travel 4 m?
time = distance / (10m/s * cos [35])

etc.

(ETA correct functions)

Also see....
http://www.glenbrook.k12.il.us/gbssci/phys/Class/vectors/u3l2f.html
Link Posted: 10/5/2005 8:58:57 PM EDT
Link Posted: 10/5/2005 9:02:28 PM EDT
You will hit the object and it will bounce back and break your nose.

Dont do it.
Link Posted: 10/5/2005 9:04:29 PM EDT
I can tell you from personal experience, the ball will land 3 feet in front of you and 9 feet to the left, but will retreat behind its previous position as it spins on the ground. There's nothing like empirical data to back up your hypothesis!
Link Posted: 10/5/2005 9:07:35 PM EDT
Link Posted: 10/5/2005 9:09:12 PM EDT
Link Posted: 10/5/2005 9:18:55 PM EDT
PE... Thanks buddy. I'm still tring to work with the wifey on figureing it all out. We have the answers already, but she and i don't understand how they got it amongst other things.

1gun... shouldn't u be playing pocket pool with TRG

Blamo... that cold weather's effecting u too much buddy... come back to TX
Link Posted: 10/5/2005 9:24:20 PM EDT

Originally Posted By Sparsky:
Need help with a problem. I'm just tring to understand how to get the answer, etc.

If I were to kick a football at 10m/sec at a 35 degree angle at an object 4 meters away and 1.5 meters high would I hit the object or over shoot it? How high or low would I be and how far would the ball travel if I miss the object? Force of gravity being 9.80m/sec squared.

If anyone knows this in "dummy" terms please LMK ASAP. Thanks

Edit to add: Weight of object unknown/not in factor along with air resistance.



The weight doesn't matter, air resistance can be assumed not to matter--in this case, due to the proximity, it pretty much doesn't.

Draw your free body diagram for the point of impact/launch. The velocity at that point is 10 m/sec at 35 degrees (angle measured from ground up to trajectory I assume). You can figure X and Y component velocities using cos and sin respectively.

Now you can construct your equations of motion. Because you are neglecting wind resistance, your football will experience no acceleration in the X direction (until it hits something anyway!). So that equation is simple, Dx=Vx*t or however you want to annotate it.

Your Y value, height, is the one that will experience acceleration--due to gravity. Choose the sign of your Y-axis (decide whether height is positive or negative). Now formulate your equation of motion in the Y direction, which will have to include the initial Y velocity,initial Y position, the Y acceleration, and time. This is simple kinematics.

Now you have an equation of motion in the X direction, and you can figure out how much time it takes for the ball to arrive at the X position of the target. You have an equation of motion in the Y direction, so you can use the TIME to figure out the Y position of the ball when it reaches the X position of the target. If the Y position of the ball is equal to the height of the target at the X position of the target, it hits. Otherwise, it misses.

For discussion, you can mention that you dont' seem to have been given the physical dimensions of the target and the ball, which means you're treating them as point objects or point positions. Realistically, for any finite target size there is a range of Y values where the ball would hit the target. I always brought this up on problems, and it always irritated the hell out of my professor...though he saw my point. :)

Jim
Link Posted: 10/5/2005 9:27:58 PM EDT
[Last Edit: 10/5/2005 9:41:35 PM EDT by roboman]
*Jumped the gun and made an asshole comment.*

Post edited.
Link Posted: 10/5/2005 9:33:23 PM EDT
[Last Edit: 10/5/2005 9:33:43 PM EDT by PromptCritical]
[Ray Finkle]

LACES UP!

[/Ray Finkle]
Link Posted: 10/5/2005 9:35:03 PM EDT
[Last Edit: 10/6/2005 9:48:19 AM EDT by BlammO]

Originally Posted By roboman:
<Edited because roboman retracted.>



Oh, for cryin' out loud.
Link Posted: 10/5/2005 9:37:11 PM EDT
[Last Edit: 10/5/2005 9:45:59 PM EDT by KS_Physicist]
(Original edited, so I'll wipe the quote, and then there's no need for my response. )
Link Posted: 10/5/2005 9:41:56 PM EDT
Edited my post out. Jumped the gun and acted like a douchebag. My apologies Sparsky.
Link Posted: 10/5/2005 9:50:45 PM EDT
[Last Edit: 10/5/2005 9:53:52 PM EDT by Sparsky]

Originally Posted By roboman:
<original comment edited out>
Edited my post out. Jumped the gun and acted like a douchebag. My apologies Sparsky.




I can understand the comment. We already have all the answers, we're just tring to figure out how to get those answers. I have only taken high school physics. My wife is the one in college, I'm tring to help her out with aid from my brothers on arfcom.

the questions i didn't post were.
What's the time at the point at which the ball is directly at ann's brother.
we have formula w/ time of .50sec

Where is the ball at Ann's bro (above/below/hit)
Above w/ .13m clearance

how long does it take for the ball to reach it's peak and total time for trip?
time .59sec total time 1.18sec

Total length traveled by ball
4.83m

Now I'm just tring to help her/me understand how they got all of this with all these lettered formulas. That and how would we get these answers with out all the answers given.

Edited to add: No worries mate.... it happens to the best of us.
Link Posted: 10/5/2005 9:57:26 PM EDT

Originally Posted By Sparsky:

Originally Posted By roboman:
<original comment edited out>
Edited my post out. Jumped the gun and acted like a douchebag. My apologies Sparsky.




I can understand the comment. We already have all the answers, we're just tring to figure out how to get those answers. I have only taken high school physics. My wife is the one in college, I'm tring to help her out with aid from my brothers on arfcom.

the questions i didn't post were.
What's the time at the point at which the ball is directly at ann's brother.
we have formula w/ time of .50sec

Where is the ball at Ann's bro (above/below/hit)
Above w/ .13m clearance

how long does it take for the ball to reach it's peak and total time for trip?
time .59sec total time 1.18sec

Total length traveled by ball
4.83m

Now I'm just tring to help her/me understand how they got all of this with all these lettered formulas. That and how would we get these answers with out all the answers given.

Edited to add: No worries mate.... it happens to the best of us.



In my example, I guided you to creating two kinematics equations--one for the X distance, and one for the Y (height) distance.

Once you have those two equations, you can compute most of what you've mentioned above. For the ones where those two kinematics equations don't quite work, you can either manipulate them directly, or you can look in the table of kinematics equations in the book to find the one that does apply.

You're given certain information in a problem, and you usually use the kinematics equation that DOES use the things you are given, and does NOT use the things you either aren't given or don't know. I assume, for instance, that you DO know where Ann's brother is standing. The X and Y kinematics equations are the tools you need to calculate the time to reach Ann's brother, the height at Ann's brother's position, etc.

Jim
Link Posted: 10/5/2005 10:17:07 PM EDT

Originally Posted By PromptCritical:
[Ray Finkle]

LACES UP!

[/Ray Finkle]



Link Posted: 10/5/2005 10:19:34 PM EDT
[Last Edit: 10/5/2005 10:21:10 PM EDT by ColonelKlink]
Link Posted: 10/5/2005 10:23:58 PM EDT

Originally Posted By PromptCritical:
[Ray Finkle]

LACES UP!

[/Ray Finkle]



I believe the correct line is "Laces Out."
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