What speed must the ball be thrown

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Homework Help Overview

The original poster attempts to determine the initial speed required for a ball to reach a maximum height of 70 meters when thrown vertically. The problem involves concepts from kinematics, particularly the effects of gravity on projectile motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using kinematic equations to relate initial velocity, acceleration due to gravity, and time. There are attempts to express the initial velocity in terms of time and gravitational acceleration.

Discussion Status

Some participants have provided guidance on using the first equation of motion and substituting values into the kinematic equations. There is ongoing exploration of different approaches to eliminate time from the equations and derive the initial velocity directly. Multiple interpretations of the problem are being examined, particularly regarding the signs of acceleration.

Contextual Notes

Participants note the importance of accounting for the negative acceleration due to gravity and question the assumptions made about the direction of motion. There is also mention of a related part of the problem that involves calculating the time of flight.

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Homework Statement



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

Homework Equations





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 + volt + (1/2)at^2, but then I need initial velocity!

What formula am I missing?
 
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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. ;)
 
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!
 
1MileCrash said:
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.
 
So Vo=gt. So replace all instances of Vo with gt in the time equation, right?

Thanks!
 
1MileCrash said:
So Vo=gt. So replace all instances of Vo with gt in the time equation, right?

Thanks!

Yes! :)
 
Do you agree with t = 2.18?
 
My answer comes out to be 3.74s when g=10 m/s^2.
It don't come 2.18s too when i take g=9.8 m/s^2
 
1MileCrash said:
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?
 
  • #10
Going that way I get vo= 21.4. Sorry I am showing no work, on a cellphone.
 
  • #11
1MileCrash said:
Going that way I get vo= 21.4.
No, but I have no idea what you did wrong.
 
  • #12
Alright, guess ill just have to wait until I get home, sit down, and think about it.
 
  • #13
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
 
  • #14
1MileCrash said:
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.
 
  • #15
Of course! Ball is thrown against gravity. Accounting for that, 3.77964473..

Thanks a lot guys.
 
  • #16
37.42 for initial velocity?
 
  • #17
1MileCrash said:
37.42 for initial velocity?

Yep you got it. :smile:
 
  • #18
1MileCrash said:
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.
 
  • #19
It wasn't unnessesary because part b asked for it anyway, but will do.
 
  • #20
1MileCrash said:
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.)
 
  • #21
1MileCrash said:

Homework Statement



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

Homework Equations


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 + volt + (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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