# Really easy velocity. why am i wrong?

1. Sep 25, 2011

### physicsgurl12

1. The problem statement, all variables and given/known data
A ball is thrown straight up with a velocity of 39m/s. how much time passes before the ball hits the ground. No air resistance.
4s-wrong
1.2s
2.4s
8s

2. Relevant equations

v=d/t a=v/t

3. The attempt at a solution
9.8/39= .02
39/9.8=3.97959

2. Sep 25, 2011

### Staff: Mentor

It would be better if you started with the general equation for vertical motion (under constant downward acceleration, like due to gravity):

$$y(t) = y_0 + v_{y0} t - \frac{1}{2} g t^2$$

If you start with that equation, you get two solutions for the time when the object is at ground level....

3. Sep 25, 2011

### DaveC426913

Well, I hope at least one of them is at t=0...

4. Sep 25, 2011

### Staff: Mentor

5. Sep 25, 2011

### physicsgurl12

okay thanks for the equation . but i solved it out and got .1 ?

6. Sep 25, 2011

### Staff: Mentor

Nope.

Please show your work, and we'll see if we can spot the error...

7. Sep 25, 2011

### physicsgurl12

okay lets see. i guess i did. 0=39m/s*t-4.9t^2
4.9/39=.1256 yeah thats not even real algebra

8. Sep 25, 2011

### Staff: Mentor

Almost there.

You should be solving for the time t, not 1/t...

Write out the algebra steps more clearly after that equation, and be sure to solve to for t.

9. Sep 25, 2011

### physicsgurl12

1.2??

10. Sep 25, 2011

### Staff: Mentor

No.

Your equation here is correct:

0=39m/s*t-4.9t^2

Just solve it for t. You came close when you tried before. Show us each algebra step. There are two terms on the righthand side of the equation. How do you move one of them to the LHS of the equation? Can you then cancel anything out?

Solve for t.

11. Sep 25, 2011

### physicsgurl12

0= 39m/s*t-4.9*t^2
4.9*t^2=39m/s*t
4.9*t=39m/s
t=7.959?

12. Sep 25, 2011

### Staff: Mentor

Bingo!

So which multiple choice answer is correct?

13. Sep 25, 2011

D. 8s yay!