Package Drop Time Calculation for Ascending Helicopter

[halbe1]

Homework Statement

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.]

Homework Equations

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

The Attempt at a Solution

I have no idea.

 

[rock.freak667]

the kinematic equation


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


will help you.

 

[halbe1]

the kinematic equation


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


will help you.

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

 

[rock.freak667]

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

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

 

[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.

 

[halbe1]

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

 

[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?

 

[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.

 

[Simply Is]

You box is traveling 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.

 

[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.

 

[halbe1]

I got 5.40 is that right?