1. Sep 21, 2009

halbe1

1. The problem statement, all variables and given/known data A helicopter is ascending vertically with a speed of 5.10 m/s. At a height of 115m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [Hint: v_0 for the package equals the speed of the helicopter.]

2. Relevant equations
x=x_0 + v_0t+ (1/2)at^2

3. The attempt at a solution
I have no idea.

Last edited: Sep 21, 2009
2. Sep 21, 2009

rock.freak667

the kinematic equation

$$y=y_0 +ut - \frac{1}{2}gt^2$$

3. Sep 21, 2009

halbe1

So what does "u" stand for and does "g" stand for gravity?

4. Sep 21, 2009

rock.freak667

u is the initial velocity and g is acceleration due to gravity

5. Sep 21, 2009

Simply Is

The "u" stands for the initial vertical velocity (or in the case of not being directly straight up... the vertical component of the initial velocity).

The "g" does stand for the gravitational constant of the Earth.

It will probably help if you just draw a cartesian grid, label your x and y, then decide where you want the positive and negative values to be.

For example, if you want to make "up" the positive y, and "right" the positive x, and consider the x-axis itself to be the Earth... Then then helicopter is traveling at a positive vertical velocity relative to that of gravity which would be in the negative direction since it is always toward Earth and you decided the x-axis was earth.

Your Ys are the displacement distances...

Plug in the values, finish the math and you are done. It's a good idea to ALWAYS draw your word problems out.

I hope that helped.

6. Sep 21, 2009

halbe1

Alright but what do I do once I get to 0=-115+5.10t+4.9t^2?

7. Sep 21, 2009

Simply Is

What are you looking for? Is what you are looking for one of the variables you have left in the equation?

If so, solve the equation.

What does the variable you show there stand for? Is that what you are looking for?

8. Sep 21, 2009

halbe1

Yea, I'm trying to get t in seconds. I tried factoring out a t and then getting two answers for t, t=0 and t=-113. So I know neither of those are right.

9. Sep 21, 2009

Simply Is

You box is travelling down... not up... and if you made toward the earth be the negative direction, then you have to put a negative direction as your final displacement... which will give you positive 115 when you move it to the other side of the equation before you plug it into the quadratic formula...

That should work... it looks like you got the initial velocity and acceleration values right.

10. Sep 21, 2009

Simply Is

Wait no... the formula is + 1/2 gt^2... and your gravity (g is basically the same for the y direction as a is in the x direction of your initial equation) is in the negative direction... so you would end up with an equation 0=115 + 5.1t - 4.9t^2...

Then you'll have the right answer.

11. Sep 21, 2009

halbe1

I got 5.40 is that right?

Last edited: Sep 21, 2009