Constant acceleration problem.

1. Sep 14, 2011

1. The problem statement, all variables and given/known data

An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.80 m away at a constant speed of 2.40 m/s, returning just in time to catch the falling ball.

1. With what minimum initial speed must she throw the ball upward to accomplish this feat?

I'm positive the answer is 23.7 m/s.

2. How high above its initial position is the ball just as she reaches the table?

2. Relevant equations

A constant acceleration equation.

3. The attempt at a solution

After I obtained the speed, I simply plugged it into the equation -9.8(1.2085)2+23.7(1.2085)

and received 14.32 m for problem number 2. Is this right? I don't feel like it is. Thanks!

Last edited: Sep 14, 2011
2. Sep 14, 2011

kuruman

How did you get 1.2085 s?

3. Sep 14, 2011

Split the time in half.

5.80/2.40 = 2.417, and then halved it.

4. Sep 14, 2011

kuruman

You halved one time too many. You forgot that she runs a total distance of 5.8+5.8 =11.6 m. So the total running time is 11.6/2.4 not 5.8/2.4.

5. Sep 14, 2011

ooh ok thanks! So I need to plug 2.417 instead of 1.2085 into the equation.

6. Sep 14, 2011

I got 33.6, sound better?

7. Sep 14, 2011

Am I using the wrong equation now? I received wrong answer for using 2.417 in the equation above.

x(t) = vt - (1/2)at^2 + x0 sounds better.

23.7(2.417)-0.5(-9.8)(2.417)^2 + 0?

xo = 0?

Last edited: Sep 14, 2011
8. Sep 14, 2011

kuruman

You have one minus sign too many. In the equation x = x0+v0t+(1/2)at2 you should replace a with -9.8 m/s2.

Yes x0 = 0.

9. Sep 14, 2011