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What speed must the ball be thrown

  • Thread starter 1MileCrash
  • Start date
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1. The problem statement, all variables and given/known data

What speed must a ball be thrown vertically to reach a max height of 70m?

2. Relevant equations



3. The attempt at a solution

I originally tried to find the velocity by the fact that the velocity of the ball must be zero at a height of 70 m, accounting for 9.8 m/s^2 acceleration due to gravity (which is negative in this case)

So, if the velocity is decreased by 9.8 m/s every second, and reaches 70 m at zero, then I could find that if I knew the time that the ball is in the air.

To find the time, I could use x=xo + vot + (1/2)at^2, but then I need initial velocity!

What formula am I missing?
 
3,810
92
Try using the first equation of motion.
v=v0+at.
You know that at the maximum height the velocity is zero, using that you can find the value of v0 (in terms of g and t).
Then substitute it in second equation of motion. ;)
 
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I'm not sure I understand how I could apply that one. I have v, as 0, but the only other value I know is a at 9.8.

0 = Vo + 9.8t

What can that tell me?

Thanks!
 
3,810
92
I'm not sure I understand how I could apply that one. I have v, as 0, but the only other value I know is a at 9.8.

0 = Vo + 9.8t

What can that tell me?

Thanks!
Oops!! You did a sign mistake :smile:
What you should get is this:-
0=v0-gt

For now, don't substitute g, when the equations are solved then you can substitute g.
 
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So Vo=gt. So replace all instances of Vo with gt in the time equation, right?

Thanks!
 
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Do you agree with t = 2.18?
 
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My answer comes out to be 3.74s when g=10 m/s^2.
It dont come 2.18s too when i take g=9.8 m/s^2
 

Doc Al

Mentor
44,733
1,024
So Vo=gt. So replace all instances of Vo with gt in the time equation, right?
Since Vo is what you want to solve for, not really the time, wouldn't it make more sense to eliminate t instead?
 
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Going that way I get vo= 21.4. Sorry I am showing no work, on a cellphone.
 
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Alright, guess ill just have to wait until I get home, sit down, and think about it.
 
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Here's my work for replacing Vo

Given that Xo = 0, X = 70, Vo = gt

[itex]x = x_{0} + v_{0}t + \frac{1}{2}gt^{2}[/itex]

Replace instances of Vo with gt, Xo is 0. x is 70.


[itex]70 = (gt)t + \frac{1}{2}gt^{2}[/itex]


[itex]70 = gt^{2} + \frac{1}{2}gt^{2}[/itex]


[itex]70 = 9.8t^{2} + 4.9t^{2}[/itex]


[itex]70 = 17.4t^{2}[/itex]

[itex]4.76 = t^{2}[/itex]

t = 2.18
 

Doc Al

Mentor
44,733
1,024
Here's my work for replacing Vo

Given that Xo = 0, X = 70, Vo = gt

[itex]x = x_{0} + v_{0}t + \frac{1}{2}gt^{2}[/itex]
Your last term has the wrong sign. The acceleration is -g, not g.
 
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Of course! Ball is thrown against gravity. Accounting for that, 3.77964473..

Thanks a lot guys.
 
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37.42 for initial velocity?
 

Doc Al

Mentor
44,733
1,024
37.42 for initial velocity?
Good. Now try doing as I suggest in post #9 and solve it without the unnecessary step of calculating the time.
 
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It wasn't unnessesary because part b asked for it anyway, but will do.
 

Doc Al

Mentor
44,733
1,024
It wasn't unnessesary because part b asked for it anyway, but will do.
It's worth doing. And in the process you'll derive a very useful kinematic formula that will enable you to solve for the velocity in one step. (Well worth committing to memory.)
 

PeterO

Homework Helper
2,425
46
1. The problem statement, all variables and given/known data

What speed must a ball be thrown vertically to reach a max height of 70m?

2. Relevant equations
3. The attempt at a solution
I originally tried to find the velocity by the fact that the velocity of the ball must be zero at a height of 70 m, accounting for 9.8 m/s^2 acceleration due to gravity (which is negative in this case)

So, if the velocity is decreased by 9.8 m/s every second, and reaches 70 m at zero, then I could find that if I knew the time that the ball is in the air.

To find the time, I could use x=xo + vot + (1/2)at^2, but then I need initial velocity!

What formula am I missing?
Now that you have the answer, I can say:

One thing to consider is that if the ball was instead dropped from 70m, it would reach the ground at the speed you sought [just down instead of up, that's why I said speed not velocity], and the time to reach the ground would have been part 2 of the question you mentioned.
 

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