mechanics question SORTED

Jun 14, 2003
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heres the question, and il try descirbe the image that comes with it as best i can

"at a given instant the football player A throws a football C with a velocity of 20m/s in the direction shown*** .determind the constant speed at which the player at B must run so that he can catch the ball at the same elevation at which it was thrown.+++ also calculate the relative velocity and relative acceleration of the football with respect to B at the instant the catch is made."

*** A throws ball at 20m/s at an angle of 60 degrees relative to the ground
therefore velocity in x = 20cos60 and in y = 20sin60

+++ player B is 15m away from A when A throws the ball

right, i have the velocities in x and y, and ive done loads of the usual SUVAT stuff to work out what the time, accel, velocity and distance/height are when the ball is in the middle of its flight.

the question does not state anyhting about friction etc, so i presume there is none, therefore making the balls journey parabolic and therefore symmetrical. correct?

the answers are in the back of the book, but for some reason they give an Angle, and the relative velocity and acceleration as answers, yet the question doesnt ask you to find an angle.

i know straight off the bat that the relative acceleration is 9.81m/s^2 as player B is not moving in the up/down direction, but the ball is.

im just completely stuck

i will try host a pic to help

edit: heres a diagram pic
 

bobsmith1492

Diamond Member
Feb 21, 2004
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Is this homework from Physics I? Looks very familiar.... ;)

Anyway, since you already have the answers, it doesn't really matter.

The first step will be to determine how long the ball will be in the air before returning to ground level. This can be done using only the acceleration due to gravity and the vertical velociity component. I'm going to assume 10 m/(s^2) for simplicity (nevermind, this requires a calculator anyway). Ok, the vertical component is 20(sin60) like you said, which is ~17.3 m/s. Divide this by the acceleration (9.81) to get the time until it stops (~1.766s). Multiply this by two to get the time until it reaches the ground again (you can prove this if you want, but it IS true - time up = time down) to get 3.5312s. Now, you can find how far the ball will travel before landing - multiply this time by the ball's velocity in the x direction, which is asumed constant: 20(cos60)(3.5312) = 35.312 m. This is where the ball will land 3.5312 seconds later. Now, the guy on the ground will have to run (35.312-15) m in 3.5312 seconds. Divide these to get 20.312m/3.512s = 5.78 m/s.

Is that right, and does it make sense??
 
Jun 14, 2003
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yes i got that, i got that the total time the ball is in the air to be 3.54 seconds, and yep i get the guy must run at 5.73m/s to catch the ball

but this is the bit that doesnt make sense

they want the relative velocity of the ball to the running man. now the horizontal component is only 10m/s for the ball, and that doesnt change since theres no acceleration in that direction once the ball is in flight. so i worked the relative velocity to be 10 - 5.73 which is 4.27m/s, but they have an answer of 17.8m/s

if the relative velocity is 17 odd, then the tru velocity must be over 20m/s how can that be since it was thrown at 20m/s, it has a parabolic path and therefore shouldnt exceed 20 at any point.

i have no idea why they state an angle in the answers, it just seems random
 

bobsmith1492

Diamond Member
Feb 21, 2004
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Hm, I'll look at it again a bit later; I've got homework now. Is that for Physics I at a university? If so, which one, and what's your major?
 

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
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Hmm.. I looked at the other part of it... when the guy catches the ball, the ball has a vertical velocity of 17.3 m/s (downward) The guy has no vertical velocity.
At that moment, the ball also has a horizontal velocity of 10 m/s. The guy has a horizontal velocity of 5.73 m/s. Relative to the guy, the ball's horizontal velocity is 4.27 m/s.

So, relative to the guy, you've got 2 components of velocity: 17.3 m/s vertically and 4.27 m/s horizontally. Pythagorean theorem.... answer is 17.8 m/s.

Oh, and they DID ask you for an angle: they asked you for the velocity relative to the receiver. By now, you should realize that 17.8 m/s is a SPEED, (or the magnitude of velocity) not a velocity, because velocity has two components: the magnitude AND THE DIRECTION. You're missing the direction. Use the vertical and horizontal components of velocity to find the direction. You're going to get an angle.
 

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
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Also, I don't know if this will help or make it harder to understand the problem...
Imagine the same scenario drawn as a cartoon. Now erase everything except the ball and the person catching the ball.

Now, we're going to watch the action on a wide-screen television... a verrrry wide screen television. All that's on the screen is the ball and the guy running. Since I'm lazy, I don't want my eyes to track the guy running across the screen, so I'm going to pay someone $5 to move the screen in the opposite direction so the guy appears to be stationary. THis is the frame of reference of the guy. When the ball hits the guy, relative to the guy, it's velocity will be about 17.8 m/s heading downward and to the right. A quick inverse tan function (no calculator in front of me at the moment) and you have the angle it's heading downward/to the right at.
 
Jun 14, 2003
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ahh DrPizza you are the man your explanation is A++++ thanks for that, i just didnt realise what exactly i was being asked, so i had all the right numbers, i was just doing the wrong thing with them. i was doing this Vb = Vc + Vb/c you know the relative velocity formula? and i got that 4.27, and i had 17.32 from earlier working. and i was just like..." these arent even close, what am i gonna do!"

cheers for that man, you've been a big help as has bobsmith1492 thankyou guys


oh n bob, im not doing physics, im studying for a masters degree in Mechanical Engieering at Loughborough University (Uk) im only in first year, so alot of the stuff we're doing is new to me. the first year is really just to bring everyone to the same point in their education, all you need do is pass the first year, as it doesnt have any bearing on your final degree score. but its nice to do well anyway.

cheers again guys