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. 

If the Government doesn't trust us with our guns, why should we trust them with theirs?

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 


"LCR, it's not for the weak; It's for the WEEKEND!"
Every penny you spend on Ebay goes to line some Commie Pinko Liberal's bank accounts . 
Posting With The Protection Of Aluminum Foil
FL, USA

You will hit the object and it will bounce back and break your nose.
Dont do it. 
"War is the remedy our enemies have chosen, and I say give them all they want."
Gen. William T. Sherman 
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!


If muslims are serious about their claims that the terrorists among them represent a minority aberration, then LET THEM CLEAN HOUSE.
Our tax dollars are funding the criminal lifestyle through the welfare system. 
Saw that in HS... Anvil, guy pulling out the object to be hit, and a BFH. = Broken nose 

"LCR, it's not for the weak; It's for the WEEKEND!"
Every penny you spend on Ebay goes to line some Commie Pinko Liberal's bank accounts . 
Saw that in HS... Anvil, guy pulling out the object to be hit, and a BFH. = Broken nose 

"LCR, it's not for the weak; It's for the WEEKEND!"
Every penny you spend on Ebay goes to line some Commie Pinko Liberal's bank accounts . 
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 

If the Government doesn't trust us with our guns, why should we trust them with theirs?

The weight doesn't matter, air resistance can be assumed not to matterin 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 accelerationdue to gravity. Choose the sign of your Yaxis (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 


*Jumped the gun and made an asshole comment.*
Post edited. 

"A large income is the best recipe for happiness I ever heard of."
J. Austin 
[Ray Finkle]
LACES UP! [/Ray Finkle] 


Oh, for cryin' out loud. 

If muslims are serious about their claims that the terrorists among them represent a minority aberration, then LET THEM CLEAN HOUSE.
Our tax dollars are funding the criminal lifestyle through the welfare system. 
(Original edited, so I'll wipe the quote, and then there's no need for my response. )



Edited my post out. Jumped the gun and acted like a douchebag. My apologies Sparsky.


"A large income is the best recipe for happiness I ever heard of."
J. Austin 
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. 

If the Government doesn't trust us with our guns, why should we trust them with theirs?

In my example, I guided you to creating two kinematics equationsone 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 




If the Government doesn't trust us with our guns, why should we trust them with theirs?

Ok kinematics can be broken down into three simple and easy to remember equations. Velocity Final = Velocity Initial + acceleration*time Vf=Vi+a*t (Velocity Final)^2= (Velocity Initial)^2 + 2*(Acceleration)*(displacement) Vf^2=Vi^2 +2*a*s Final Displacement= Initial Displacement + Initial Velocity* Time + 1/2*acceleration*(time)^2 Sf= Si + Vi*t + 1/2*a*t^2 So. In this case, what are we trying to accomplish? We want to know if the ball is ever at x=4, y=1.. So we use our third equation, but we first need to break things down. Now we have a vector, it says that the ball travels at a velocity of 10 m/s, at an angle of 35 degrees from the horizontal. We want to break this down into x and y components. We must fall back on our knowledge of trigonometry. The cosine function is cos(angle)= (adjacent leg of angle) / (hypotenuse of angle). So call the angle the greek letter theta, cos(theta)=adjacent/hyponetenuse. We know the hypotenuse, which is 10 m/s, we use algebra to solve for the adjacent leg. So the horizontal velocity is the adjacent leg and it is equal to 10 m/s * cos(35), which is about 8.2m/s. Sine(theta)=opposite/adjacent so y velocity is 5.73. So intial velocity in the x direction is about 8.2 m/s. Now that we know this we can plug it into our third equation. We want to know which t satistifies the 4 meter horizontal condition. Remember there is no acceleration so that gets rid of the squared term, and no initial displacement So, 4meters= 0 meters + 8.2*t +.5*0*t^2 4m=8.19m/s*t meters cancel out, leaving only seconds t=.49 seconds This means that the ball can only be 4 meters away at roughly half a second. We need to know where it is hieght wise at that time. Use the third equation again y= 0 + 5.73*(.49)+.5*9.8*(.49)^2= 1.63 This says that at .49 seconds the ball is 1.63 meters high, so it misses the target by .13 meters. The ball reaches its peak when vertical velocity is zero, 0=5.739.8*t 5.73=9.8 t=.59 so at t=.59 seconds the ball is at its peak height. and to find the total distance traveled you simply multiply total flight time by velocity.(this is a seperate computation, it is easy but im too tired and im going to bed) 

[15:20] <SubnetMask> Oh YEAH?!!! Well...Well...as soon as I get those better magazines, smooth the rails, change the sights, and replace the extractor...THEN we'll see who's laughing, there buddy boy.

I believe the correct line is "Laces Out." 

"A large income is the best recipe for happiness I ever heard of."
J. Austin 
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