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Throwing a Ball...

by VU2
Tags: ball, throwing
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VU2
#1
Feb10-13, 10:30 PM
P: 35
Q1) You throw a ball. Assume that the origin is on the ground, with +y-axis pointing upward. Just after the ball leaves your hand its positon is <.06,1.03,0>m. The average velocity of the ball over the next .7s is <17,4,6>m/s. At time .7s after the ball leaves your hand, what is the height of the ball above the ground.

y=1.03+4(.7)-9.8/2(.7^2)
y=1.429 is my answer, is this correct? I'm using the average velocity as initial, so i'm not sure if I can assume that or not.
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Simon Bridge
#2
Feb10-13, 10:59 PM
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Welcome to PF;
I'm using the average velocity as initial, so i'm not sure if I can assume that or not.
Probably not.
Check it to see if it matters. Depends how accurate your answer needs to be.
You know the acceleration and the average velocity, can you find the y-component of the initial instantaneous velocity.
Note: how is "average velocity" calculated?
VU2
#3
Feb11-13, 01:25 AM
P: 35
AvgV is calculated by (finalV+initalV)/2=AvgV, can i assume that initial V is 0?

tms
#4
Feb11-13, 01:33 AM
P: 616
Throwing a Ball...

No, you can't assume that. You can, however, calculate the final velocity in terms of the initial velocity, given the acceleration and the time. That will allow you to get the initial velocity from the average.
VU2
#5
Feb11-13, 01:39 AM
P: 35
Oh I see, y-final=y-initial + v(avg)*(t). Therefore, y-final=1.03+4(.7)=3.83?
tms
#6
Feb11-13, 01:57 AM
P: 616
That will work, too, more directly than what I suggested. I don't know what the notation "<17,4,6>" means, so I don't know if your answer is right.
VU2
#7
Feb11-13, 02:18 AM
P: 35
That is the average velocity vector.
VU2
#8
Feb11-13, 02:19 AM
P: 35
I'm curious now, how would you compute it your way?
tms
#9
Feb11-13, 02:32 AM
P: 616
Quote Quote by VU2 View Post
That is the average velocity vector.
That's what I thought, but I was wondering why the z direction was there, since it isn't mentioned in the rest of the problem.
tms
#10
Feb11-13, 02:33 AM
P: 616
Quote Quote by VU2 View Post
I'm curious now, how would you compute it your way?
I explained it above. It is essentially the same, except that it takes a roundabout way to get there (basically deriving the equation you used).
VU2
#11
Feb11-13, 02:40 AM
P: 35
Oh I see, thanks!
Simon Bridge
#12
Feb11-13, 06:46 AM
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AvgV is calculated by (finalV+initalV)/2=AvgV, can i assume that initial V is 0?
Average velocity is change in position over change in time:
##\bar{v}=\Delta y/\Delta t = (y_f-y_i)/\Delta t##
... you rearranged that equation to give you the final height given the initial height and the average velocity ... well done.

As an exercise - how would it have been different for the same figures, except the average velocity was timed over 0.1s, but you still want the final height after 0.7s?
VU2
#13
Feb14-13, 05:09 PM
P: 35
Simon Bridge,

Sorry, I was suppose to get back to you sooner. That's an interesting question. I'm not sure how I would approach it, to be honest.
Simon Bridge
#14
Feb15-13, 06:19 AM
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Well - in that case you'd have two time periods ot consider: one short one (with a displacement) to establish the initial velocity and another, longer, one to find the answer. Otherwise the approach is identical to the one you used above.

IRL you often measure the speed of something by timing it over a short distance, and then use that information to work out where the thing will end up.
VU2
#15
Feb15-13, 12:43 PM
P: 35
Simon Bridge,

Yeah that's what I figured too. Thanks for the lesson.
Simon Bridge
#16
Feb15-13, 09:18 PM
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No worries - have fun :D


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