[PHY]Simple Kinematic Question

[AbtLm]

Dealing with kinematic motion problems is one of my weaknesses. It would be great if someone can help me solve this simple problem:

"A ball is thrown vertically with 3 m/s. What's the ball's total time in flight if it has traveled 50cm?"

Thanks.

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[apoptos]

Dealing with kinematic motion problems is one of my weaknesses. It would be great if someone can help me solve this simple problem:

"A ball is thrown vertically with 3 m/s. What's the ball's total time in flight if it has traveled 50cm?"
Thanks.

I think because it's only vertical, you can do this without the distance, but someone should check my reasoning here.

final v= initial v + at
the final v should be zero as it reaches its maximum height,
0 = 3m/s + 10 m/s^2 (t)
-3 = 10t, 3/10 = t.

that's the trip up...but since it comes back down, you want to double t, so .6 seconds.
Is that right...?

 

[Psychotropic]

Dealing with kinematic motion problems is one of my weaknesses. It would be great if someone can help me solve this simple problem:

"A ball is thrown vertically with 3 m/s. What's the ball's total time in flight if it has traveled 50cm?"

Thanks.

Use: delta Y= .5*a*t^2 +v(initial)*t

 

[J ROD]

I think because it's only vertical, you can do this without the distance, but someone should check my reasoning here.

final v= initial v + at
the final v should be zero as it reaches its maximum height,
0 = 3m/s + 10 m/s^2 (t)
-3 = 10t, 3/10 = t.

that's the trip up...but since it comes back down, you want to double t, so .6 seconds.
Is that right...?

yeah, you are right. This way is much faster to me than the equation above.

As long as you know the max vertical velocity is zero.

 

[AbtLm]

To apoptos: I initially thought so, too, but the answer key says that the total time flight is 0.398 sec.

Any ideas how the key came to this value? Thanks, guys. Or maybe the key's wrong.

 

[apoptos]

...but the answer key says that the total time flight is 0.398 sec...maybe the key's wrong.

Hmm... I hope so. Otherwise, I'm stumped.
What's the source of the question? Maybe we have the same book...

 

[smeagol]

As previously stated, it takes t = 0.3 s to reach its maximum height.

Plug that into the equation y = 3t - 5t^2 and you get .45 m

The question says it travels another .05 m so:

Since it just falls .05 m after in flight for .3 s:

0.05 = 5t^2

t = 0.1

Total time = 0.3 + 0.1 = ~0.4 s

What is confusing is why it stops after .05 m of falling...maybe someone caught it?

 

[apoptos]

As previously stated, it takes t = 0.3 s to reach its maximum height.

Plug that into the equation y = 3t - 5t^2 and you get .45 m

The question says it travels another .05 m so:

Since it just falls .05 m after in flight for .3 s:

0.05 = 5t^2

t = 0.1

Total time = 0.3 + 0.1 = ~0.4 s

What is confusing is why it stops after .05 m of falling...maybe someone caught it?

I think you did. I mean, NICE CATCH!

 

[AbtLm]

apoptos: It's from KAPLAN's high-yield problem solving guide.

smeagol: Thank you for the clarification. Apparently, I need more practice with this type of question.

 

[drkryptonight]

wait so whats right? i know how to find the flight time if the ball is thrown horizontal... just not the thrown vertical part. am i ******ed?

 

[Arc]

Dealing with kinematic motion problems is one of my weaknesses. It would be great if someone can help me solve this simple problem:

"A ball is thrown vertically with 3 m/s. What's the ball's total time in flight if it has traveled 50cm?"

Thanks.

You gotta use a kinematic Formula. So solve for time using .50M = vt+1/2at^2
(I already converted the cm to M for you) This problem is cake...

 

[Arc]

wait so whats right? i know how to find the flight time if the ball is thrown horizontal... just not the thrown vertical part. am i ******ed?

No , you just don't know the ins and outs of the formulas. When the ball is thrown horizontal, It will go up to a certian height, hits a transition point, and fall downwards at the speed of gravity. (Hence the 1/2at^2 part)

 

[smeagol]

You gotta use a kinematic Formula. So solve for time using .50M = vt+1/2at^2
(I already converted the cm to M for you) This problem is cake...

It's really not that simple. You might be misleading people if you claim that equation as explicitly described above gives you the answer for t.

 

[Arc]

It's really not that simple. You might be misleading people if you claim that equation as explicitly described above gives you the answer for t.

I forgot to mention that those t's are really change of time at that point. Is that what you are inferring to?

 

[smeagol]

.50M = vt+1/2at^2

Eck I was hoping you would have just read the posts above.

That equation gives you a nonexisting answer... (unless you meant -.5 m = vt + 1/2at^2, but is still incorrect)

More importantly is the fact that the position vector is never .5 meters; its maximum height is .45 meters (.5 m is total distance traveled, meaning that someone stopped it before it could completely fall back down). So setting the position equation = .5 m doesn't make a lot of sense.