Catching a Falling Can: Solving for Acceleration

[Jeramaya]

Homework Statement

"A friend leans out of an upper window of an apartment building and drops a canned beverage of your choice to you on the sidewalk below. Unfortunately, you aren't directly beneath the window; instead, you are standing where the angle to your friend is 75 degrees above horizontal. Thinking quickly, you start accelerating toward the point where the falling can will reach ground level. Find the constant acceleration of your motion that would allow you to catch the can as you arrive."

Homework Equations

The Attempt at a Solution

What I know is that I made up a weight for the can (.75 lbs), which I converted to Newtons (.75lb x 1N/.2248lbs = 3.3N). I know that I weigh 140 lbs, which I converted to 623 Newtons. The falling can has the equation F=ma, so I calculated the mass to be .34kg. The final y-position is zero--the point where I catch the can. The initial velocity in the y-direction for the can, as well as the initial velocity in the x-direction for me, are both zero. Since there is no time or final velocity given, I am really confused as to how to approach this problem. Any help/hints are greatly appreciated.

 

[timmay]

1. It will probably help you if you draw a diagram of the problem.

2. The acceleration of a falling object, neglecting resistances, is independent of weight. You should look at the basic equations of kinematics, found here.

3. Essentially, the catcher will have to reach the point where the object will hit the ground (ignore the fact that it would be easier to catch at head height, it's probably not taken into consideration for the question) at the same time as the falling object does. So you will have to calculate what distance the catcher is away from that point, and how long the object will take to fall to the ground.

Have a go and see how you do.

 

[Carid]

Don't worry about the mass of the can. Galileo Galilei demonstrated a long time ago that a small connon ball and a big cannon ball dropped from the top of the tower in Pisa both reached the ground at the same moment. Both cannon balls accelerated at g = 9.8metres/sec/sec.
The point is to think about the shape of the triangle linking you, your friend and the spot where you will catch the can. The ratio of the length of the vertical side of the triangle and the length of the horizontal side of the triangle will be numerically the same as the ratio of the acceleration of the can and the acceleration you'll have to make. No forces need to be calculated.

 

[Jeramaya]

OK. So if I take the bottom of the triangle to be side A, and the vertical side of the triangle to be B, I can set it up like this:

Tan 75 degrees = B/A = Accel. of can in y-direction / Accel. of me in x-direction

Then,

Accel. of me in x-direction = Accel. of can in y-direction / Tan 75 degrees
= 9.80m/s/s / Tan 75 degrees = 2.63 m/s/s

Does my mass have nothing to do with the equation then?

 

[Carid]

No. Forget mass.

The equation to consider is one of the kinematic ones for constant acceleration

DistanceCovered = (InitialVelocity*Time) + 0.5*Acceleration*Time*Time

For you and the can, the InitialVelocity is zero, so we can neglect that part of the equation.

For both you and the can, the Time taken is the same. (Otherwise you'll be too early or too late to catch it).

So we find that for both you and the can, the DistanceCovered is proportional to the Acceleration.

I agree with your numerical result.