# What speed must the ball be thrown

1. Aug 24, 2011

### 1MileCrash

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?

2. Aug 24, 2011

### Saitama

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. ;)

3. Aug 24, 2011

### 1MileCrash

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!

4. Aug 24, 2011

### Saitama

Oops!! You did a sign mistake
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.

5. Aug 24, 2011

### 1MileCrash

So Vo=gt. So replace all instances of Vo with gt in the time equation, right?

Thanks!

6. Aug 24, 2011

### Saitama

Yes! :)

7. Aug 24, 2011

### 1MileCrash

Do you agree with t = 2.18?

8. Aug 24, 2011

### Saitama

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

9. Aug 24, 2011

### Staff: Mentor

Since Vo is what you want to solve for, not really the time, wouldn't it make more sense to eliminate t instead?

10. Aug 24, 2011

### 1MileCrash

Going that way I get vo= 21.4. Sorry I am showing no work, on a cellphone.

11. Aug 24, 2011

### Staff: Mentor

No, but I have no idea what you did wrong.

12. Aug 24, 2011

### 1MileCrash

Alright, guess ill just have to wait until I get home, sit down, and think about it.

13. Aug 24, 2011

### 1MileCrash

Here's my work for replacing Vo

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

$x = x_{0} + v_{0}t + \frac{1}{2}gt^{2}$

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

$70 = (gt)t + \frac{1}{2}gt^{2}$

$70 = gt^{2} + \frac{1}{2}gt^{2}$

$70 = 9.8t^{2} + 4.9t^{2}$

$70 = 17.4t^{2}$

$4.76 = t^{2}$

t = 2.18

14. Aug 24, 2011

### Staff: Mentor

Your last term has the wrong sign. The acceleration is -g, not g.

15. Aug 24, 2011

### 1MileCrash

Of course! Ball is thrown against gravity. Accounting for that, 3.77964473..

Thanks a lot guys.

16. Aug 24, 2011

### 1MileCrash

37.42 for initial velocity?

17. Aug 25, 2011

### Saitama

Yep you got it.

18. Aug 25, 2011

### Staff: Mentor

Good. Now try doing as I suggest in post #9 and solve it without the unnecessary step of calculating the time.

19. Aug 25, 2011

### 1MileCrash

It wasn't unnessesary because part b asked for it anyway, but will do.

20. Aug 25, 2011

### Staff: Mentor

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.)