# Easy free-falling kinematics question help!

1. Jul 8, 2010

### elasticities

1. The problem statement, all variables and given/known data
If the rock is thrown from the top of the building with a velocity of 5m/s [Up].
And you know acceleration due to gravity is 9.81m/s^2 [Down].
And the height of the building is 50m.

Please help! I can't find the time for it to reach the bottom! I am letting up be positive!

2. Relevant equations
d = V1(t) +1/2(a)(t^2)

That's the only one I can think of! Help!

3. The attempt at a solution
The only way I can do it is to plug in the givens into the formula, am I doing it right or is there something more?

-50=5(t)+1/2(-9.81)(t^2)
0=-4.90500(t^2)+5t+50
t=3.74s or t=-2.73s

2. Jul 8, 2010

### Karmalo

It is right. Time needs to be positive because we need a result that is after we started the experiment. t=-2.73s is negative, what means it is before the beginning of the experiment so the only correct answer is t=3,74s

3. Jul 8, 2010

### Staff: Mentor

Looks OK to me. Which of those times do you think is the right one? (Only one makes physical sense for this problem.)

4. Jul 8, 2010

### elasticities

Thanks for your input, could you see my concern below (as a reply to Doc Al), too?

The positive time of course. But my issue was that I don't know if I seem to have included the time during which it went upwards reached its max height and then went down. Or is that included in the equation because I set specific signs for directions? Please let me know if you need more explanation of my doubt.

5. Jul 8, 2010

### Staff: Mentor

That equation takes everything into account. The initial rise to a max height, then the fall to the ground. (Just for fun, see if you can figure out how long it takes to get to the maximum height.)

6. Jul 8, 2010

### elasticities

Thank you so much for your help! You have no idea how much I appreciate it! :)

7. Jul 8, 2010

### pgardn

Neat thing about this is, if you had thrown the rock straight up from the ground, 2.73 seconds later the rock would be 50 m above the ground moving at 5 m/s up. Then 3.74 seconds later from this, the rock would strike the ground a the same velocity it was thrown up at. And of course you could find out the velocity it hits the ground at.

So the math does have some meaning, but like Doc Al said, not in the context of the limits given in the problem.