Question

Jamie decided to drop an egg that has a mass of 345 g off the top of a 6.7 m building. How fast will it fall right before it hits the ground?

Answer 1

Assuming the egg is in free fall and air resistance is negligible, we can draw a free body diagram for the egg. Do you know how to do this?

Answer 2

I drew out the building and the egg falling with the mass and the height, if thats what you mean

Answer 3

No by free body diagram i mean a force diagram

Answer 4

Oh, then no

Answer 5

There might be a different way to solve this and I am going to use that way over the way I was going to do We can use vertical kinematics and just ignore the mass given to us

Answer 6

Our unknown is how fast it is falling at the end correct?

Answer 7

Yep :)

Answer 8

Well what do we know. Its mass which we can ignore. How high it is which is 6.7m, its initial y-velocity of 0 m/s, and the acceleration which is g or 9.8 m/s^2. We can not simply find the final velocity from just this info, we need one more piece of information the time it takes to fall. We have an equation for this if we know g, initial velocity is 0, and a height..

\[y = \frac{ 1 }{ 2 }g t^{2}+v_{yo}t+y_{o}\]

Right?

Answer 9

Yeah :)

Answer 10

Actually I am revising once more lol there is an even easier equation we can use without time. Does this equation look familiar to you?

\[v_{y}^2=v_{oy}^{2}-2gy\]

Answer 11

Haha yes, vaguely. I think my teacher used it once but I can't remember, it was awhile ago if he did

Answer 12

Well this eliminates our time and we can substitute our height in for y, 9.8 for g and 0 for initial velocity and solve for final velocity. This you can do that?

Answer 13

You will end up not being able to squareroot a negative just ignore the negative and find the squareroot and add the negative back on since the negative is merely for direction since velocity is a vector quantity and downwards is negative

Answer 14

I got -1.037 but i dont think it's right? I probably did the math wrong.

Answer 15

should get 11 something. lets take it one step at a time. plug our values in first

\[v_{y}^{2}=0-2\left( 4.9m/2^2 \right)\left( 6.7m \right)\]
\[v_{y}=-\sqrt{\left( 4.9m/2^2 \right)}\left( 6.7m \right)\]
\[v_{y}=-\sqrt{131.32m^2/s^2}\]
\[v_{y}=-11.46m/s\]

Answer 16

Ohhh I see what i did. I forgot to square it first. Thank you!!!

Answer 17

your welcome :)

Answer 18

aso realized my seconds were 2's by accident oh well

Answer 19

Haha i understood what you meant :D

Answer 20

Good

Answer 21

Understand that first method we were about to do would of worked also