Free fall physics word problem

Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his broefcase when he is 130m above the water. If the briefcase hits the wate 6.0seconds later, what was the speed at which the helicopter was ascending?

The book says the answer is 7.7m/s but I got like 50.

Answer 1

Alex,

 

I confirm the book answer and the answers provided by Steven and Karin.  I think I figured out the mistake that led to a value of about 50:

 

h(t) = h₀ + v₀t - (1/2)gt²

 

At t = 6s, h(t) = h(6) = 0m

 

Then at t = 6s,

 

0 = h₀ + v₀t - (1/2)gt²

 

v₀ = [(1/2)gt² - h₀] / t = [(1/2)(9.8)(6²) - 130] / 6 ≅ 7.7 m/s

 

I suspect you added h₀ in the equation above, rather than subtracting it.  Note that adding h₀ gives

 

[(1/2)(9.8)(6²) + 130] / 6 ≅ 51, which is about what you got.  Please check your work and see if this is so.

Answer 2

Hi Alex!

 

Let's put this problem in some context, so you can also solve others like it.

 

Kinematics, the subject of this problem, is the study of motion using five key quantities: acceleration (a), initial velocity (vo), final velocity (v), displacement (x-xo, where x is final position and xo is initial position), and time (t). Usually, a problem asks you to solve for one of these quantities. You should have four equations at your disposal, relating these five quantities. Each equation involves four of the five quantities. Therefore, if you want to use the equations to solve for one of the kinematic quantities, you have to know at least three others for the given situation. The challenge of kinematic problems is usually finding out what information you are given.

The way I usually work these problems is to ask, first, which of the five kinematic quantities is being asked for? This is often easiest to pick out, because the question has to directly state it. In this problem, we are asked about the initial velocity of the helicopter.  Now, the briefcase is the object in kinematic free fall here, so we need to answer a question about IT.  But since it was traveling with the helicopter at the start, its initial velocity is the helicopter's initial velocity.  So, we can say that we are trying to solve for the briefcase's initial velocity:

 

to find:  v_o

 

To solve for this, we have to know three other kinematic quantities among the remaining four -- a, v, t, and (x-x_o) -- for the interval where the briefcase falls from the villain's hand to the water.  We have to scrub the rest of the problem to figure which of these we know.

 

For any object is free fall (meaning air resistance is being ignored), we immediately know it is traveling only under the acceleration of gravity.  So, a = -g  (I assume downward means negative).

 

We are told the time of its fall is t = 6 s, so that is a second kinematic quantity we know in this case.

 

Finally, we are told the briefcase started at 130 m above the water and then fell to the water.  So it moved down by 130 m, making its displacement -130 m (since down is negative).

 

So we can list what we know:

 

know:  a, t, (x-x_o)

 

a = -9.8 m/s²

t = 6 s

(x-x_o) = -130 m

 

Now, we have to go to our stable of kinematic equations (usually there are four) and choose the one which includes the three quantities we know, and the one we want to find.  In this case, that would be:

 

(x-x_o) = v_ot+(1/2)at²

 

We can put in what we know and solve for what we want to find:

 

-130 m = v_o(6 s)+(1/2)(-9.8 m/s²)(6 s)²

 

This can then be solved for v_o.  I confirm that it works out to v_o = (+) 7.7 m/s.

 

This procedure will work no matter which three quantities you know and which you want to find (provided one of those four is initial velocity), and you can use ANY of the kinematic equations to help you solve a lot of these in one step.

 

I hope this helps!

Answer 3

You need to use the distance formula being careful about vectors and direction. 

 

df=d_i+v₀t+(1/2)at²

remember the velocity it up and accelation is down so pay attention to the signs. 

 

0=130+v₀(6)-(1/2)(9.8)(6)²

0=130+v₀(6)-176.4

46.4=v₀(6)

so the speed is 7.7 m/s