Calculating the Final Velocity of a Thrown Ball

[Medgirl314]

Homework Statement

A boy throws a baseball directly upward. He catches it 3.2 seconds later.

Homework Equations

a=Δv/Δt

The Attempt at a Solution

Letting up be positive, I think the acceleration would be positive 9.8 m/s^2. I think I am trying to find the final velocity. I think that the inital velocity would be 0, right before he throws the ball. Could I go about this problem by using a=Δv/Δt to find velocity? This method seems to yield an answer of 3.0625 m/s. Thanks in advance for a helpful reply!

 

[berkeman]

Homework Statement

A boy throws a baseball directly upward. He catches it 3.2 seconds later.

Homework Equations

a=Δv/Δt

The Attempt at a Solution

Letting up be positive, I think the acceleration would be positive 9.8 m/s^2. I think I am trying to find the final velocity. I think that the inital velocity would be 0, right before he throws the ball. Could I go about this problem by using a=Δv/Δt to find velocity? This method seems to yield an answer of 3.0625 m/s. Thanks in advance for a helpful reply!

If up is positive, gravity's acceleration is downward, so it is negative.

What is the equation for the vertical position of the ball versus time? Just solve for Vi given that the ball returns to the same y position after 3.2 seconds.

 

[Medgirl314]

If up is positive, gravity's acceleration is downward, so it is negative.

What is the equation for the vertical position of the ball versus time? Just solve for Vi given that the ball returns to the same y position after 3.2 seconds.

Thanks berkeman! Are you saying that I should use this equation, but take into account that the acceleration is negative?

 

[berkeman]

Thanks berkeman! Are you saying that I should use this equation, but take into account that the acceleration is negative?

Which what where equation?

 

[Medgirl314]

Sorry, wrong demonstrative pronoun. *That* equation.

Which is *this* equation:a=Δv/Δt.

 

[berkeman]

Sorry, wrong demonstrative pronoun. *That* equation.

Which is *this* equation:a=Δv/Δt.

Oh, now we're using BIG words!

No, I mentioned the equation for the vertical position of the ball as a function of time. Can you start with that equation?

 

[HallsofIvy]

Try using s(t)= -at^2+ vt+ h where a= 9.8 m/s^2, v is the initial velocity and h the initial height. Since you are talking about a ball going up and back down, you can take the initial height to be 0. The quadratic equation, -9.8t^2+ vt=0, has two solutions. One is obviously 0, the time the ball leaves your hand. The other solution is the time the ball returns to your hand which you are told 3.2 seconds. So solve -9.8(3.2)^2+ v(3.2)= 0 for v.

 

[Medgirl314]

Haha! You should stick around.

I may be able to, but I haven't gotten into vertical position yet, unless you're talking about the change in height.

 

[Medgirl314]

Try using s(t)= -at^2+ vt+ h where a= 9.8 m/s^2, v is the initial velocity and h the initial height. Since you are talking about a ball going up and back down, you can take the initial height to be 0. The quadratic equation, -9.8t^2+ vt=0, has two solutions. One is obviously 0, the time the ball leaves your hand. The other solution is the time the ball returns to your hand which you are told 3.2 seconds. So solve -9.8(3.2)^2+ v(3.2)= 0 for v.

Thank you! I will try this and post my answer in a moment.

 

[Medgirl314]

983.4496+v(3.2)=0

Do I take the square root of both sides now? Or do I use the quadriatic equation?

 

[Medgirl314]

I think I can use the quadriatic equation, but I'm having some trouble figuring out which numbers to assign to the letters.

 

[berkeman]

I think I can use the quadriatic equation, but I'm having some trouble figuring out which numbers to assign to the letters.

Please show more of your work. I know that Halls basically solved the problem for you, but you need to do some of this schoolwork on your own.

Show us the definition of the Quadratic Equation. Show us how it works (look at wikipedia, for example). Then show us how you think you should use it to solve this problem.

 

[Medgirl314]

Well, I wouldn't have had trouble with that part anyway, I supposse, it's just assigning the letters that seems difficult. Here's the definition I'm looking at: http://www.purplemath.com/modules/quadform.htm-9.8(3.2)^2+ v(3.2)= 0

Would -9.8 be a, v would be b, and 3.2 would be c?

Thanks!

 

[Medgirl314]

Or would -9.8(3.2) be a, v would be b, and 3.2 would be c?

 

[berkeman]

Actually, since the equation does not have a constant term (since the ball returns to the same height), you can just solve the equation directly for Vi. Sorry that I missed seeing that earlier. What answer do you get?

 

[Medgirl314]

I'm not quite sure yet. Do I take this:-9.8(3.2)^2+ v(3.2)= 0 and divide by v? Or subtract v?

 

[berkeman]

I'm not quite sure yet. Do I take this:-9.8(3.2)^2+ v(3.2)= 0 and divide by v? Or subtract v?

Start with that equation and take the first term on the lefthand side (LHS) over to the RHS...

 

[Medgirl314]

Okay, thanks! I've done this plently of times with binomials and trinomials, but the quadriatics seem a bit harder.
-9.8(3.2)^2+ v(3.2)= 0
v(3.2)+9.8(3.2)=0

Like that?

 

[berkeman]

Okay, thanks! I've done this plently of times with binomials and trinomials, but the quadriatics seem a bit harder.
-9.8(3.2)^2+ v(3.2)= 0
v(3.2)+9.8(3.2)=0

Like that?

I'm not following what you did. My suggestion was to take the first term on the LHS over to the RHS. That would put the = sign between the two terms. Then simplify...

 

[Medgirl314]

I added the -9.8(3.2)^2 to v(3.2). Oops.

-9.8(3.2)^2+ v(3.2)= 0
v(3.2)=9.8(3.2)

More like that?

 

[berkeman]

I added the -9.8(3.2)^2 to v(3.2). Oops.

-9.8(3.2)^2+ v(3.2)= 0
v(3.2)=9.8(3.2)

More like that?

It looks like you dropped the ^2 part on the RHS, but that's along the right lines other than that...

 

[Medgirl314]

Oops, typo. Thanks! Where did the =0 go?
v(3.2)=9.8(3.2)^2
v(3.2)=885.6576
276.768=885.6576

That doesn't add up. Where did I go wrong?

 

[berkeman]

Oops, typo. Thanks! Where did the =0 go?
v(3.2)=9.8(3.2)^2
v(3.2)=885.6576
276.768=885.6576

That doesn't add up. Where did I go wrong?

Something is wrong with your calculator...

On the RHS, square the 3.2, *then* multiply by 9.8.

Then divide the RHS by 3.2 to solve for Vi...

 

[Medgirl314]

Okay, thanks! I thought the whole thing was squared.

v(3.2)=9.8(3.2)^2
v(3.2)=9.8(10.24)
v(3.2)=100.352
v=31.36 m/s

Is that my final answer?

 

[berkeman]

Okay, thanks! I thought the whole thing was squared.

v(3.2)=9.8(3.2)^2
v(3.2)=9.8(10.24)
v(3.2)=100.352
v=31.36 m/s

Is that my final answer?

Looks reasonable.

Remember, the equation you were starting with was:

$$y(t) = y_0 + v_{iy} * t + \frac{1}{2} * a_y * t^2$$

It's important to keep your terms in mind, so that you don't end up squaring the wrong thing(s).

 

[Medgirl314]

Thank you! Now that I have the inital velocity, and the final velocity is 0(I assume) I can use vavg=distance/time to find height, correct?

 

[berkeman]

Homework Statement

A boy throws a baseball directly upward. He catches it 3.2 seconds later.

Homework Equations

a=Δv/Δt

The Attempt at a Solution

Letting up be positive, I think the acceleration would be positive 9.8 m/s^2. I think I am trying to find the final velocity. I think that the inital velocity would be 0, right before he throws the ball. Could I go about this problem by using a=Δv/Δt to find velocity? This method seems to yield an answer of 3.0625 m/s. Thanks in advance for a helpful reply!

Thank you! Now that I have the inital velocity, and the final velocity is 0(I assume) I can use vavg=distance/time to find height, correct?

I thought the question was what does Vi have to be so that you catch the ball 3.2s after throwing it straight up. Is there more to the question? And no, when you catch the ball it is traveling downward with velocity -Vi. Can you see why that is?

 

[Medgirl314]

Sorry, that was the question. It had two parts, but I hoped after I found the first answer I could easily get the second. The second question is "How high does the ball go?" So can I use that equation as long as I say vi is negative? Thanks!

 

[berkeman]

Sorry, that was the question. It had two parts, but I hoped after I found the first answer I could easily get the second. The second question is "How high does the ball go?" So can I use that equation as long as I say vi is negative? Thanks!

If it takes 3.2 seconds for the ball to go up and come back down, what is the time when the ball is at the top of its travel? Use that time and the distance equation...

 

[Medgirl314]

Great, thanks! The equation I used, or a different one?

 

[berkeman]

Great, thanks! The equation I used, or a different one?

The distance versus time equation that I re-summarized in post #25...

 

[Medgirl314]

Okay, I see it now. Thanks! I'll post my answer soon.

 

[Medgirl314]

y(t)=y0+viy∗t+(0.5∗ay)∗t2
y(3.2s)=0+0*3.2+(0.5)(-9.8y)*10.24
y(3.2s)=3.2s+(0.5)(9.8y)*10.24sIs that right so far? What's next?

 

[berkeman]

y(t)=y0+viy∗t+(0.5∗ay)∗t2
y(3.2s)=0+0*3.2+(0.5)(-9.8y)*10.24
y(3.2s)=3.2s+(0.5)(9.8y)*10.24s


Is that right so far? What's next?

I have no idea what you wrote. Could you please narrate what you are trying to do?

 

[Medgirl314]

I'm not really sure. I was trying to plug in everything I know into the equation from post #25. It went horribbly wrong.