Homework Help: Gravity Question

1. Oct 11, 2006

XPX1

Seagulls are often observed dropping clams and other shellfish from a height to the rocks below, as a means of opening the shells. If a seagull drops a shell from rest at a height of 18 m, how fast is the shell moving when it hits the rocks?

It seems as though I can figure out some steps of a problem, but when it comes to the rest of them, I am completly lost.

Since the height is 18 feet, the initial velocity(vi) is 0, and the final velocity(vf) is 18

Gravity is 9.8 or -9.8, so I know the acceleration, so I use the equation

vf=vi+at

18=0+(9.8)t
18=9.8t
18/9.8=1.836734694 or 1.84
t=1.84

vi=0
vf=18
a=9.8(gravity) or -9.8 since its moving down?
t=1.84

Now that I know the following, It takes 1.84 seconds for a seagull to drop a clam from a height of 18m.

Do I multiply 1.84 by 18 to get how fast the clam is moving when it hits the rocks?

*Also, if there is any website out there that helps explain some problems in greater depth, or how to develop a better thought process, I dont know, please be sure to include a link!

2. Oct 11, 2006

You are asked to find its final velocity just before it hits the ground. You have incorrectly stated it as vf =18m/s. 18m is the distance it travels, not its speed. Why did you assume the final velocity was 18m/s? Why not use the motion equation that relates displacement with velocity $$v_f^2 = v_i^2 + 2gh[tex]? 3. Oct 11, 2006 BishopUser why are you assuming Vf = 18? vf is what you are trying to calculate! You should not use that equation because you don't know vf and you don't know time. Try using the equation... [tex]V_f^2 = V_i^2 + 2a(\Delta y)$$

4. Oct 11, 2006

XPX1

Thanks for pointing that out, I assumed that velocity was 18 because, it started at 18.. Duh, the initial velocity is 18! So if I plug it into the equation

Vf^2=Vi^2+2a(xf-xi)

vf^2=18^2+2(-9.8)(18-0)

The Final Position xf=18, because wherever it is, it has to go 18 meters down.

The Initial Position x=0 because wherever it is, it had to start from 0?

So vf^2=-28.8?

And then to get vf I would take the square root of -28.8? But how can I do that if it is negative, wouldnt I get an imaginary number?

I also put gravity as -9.8 because it is going down.

5. Oct 11, 2006

BishopUser

where are you getting this idea that that 18 has anything to do with velocity. Its initial velocity will be 0 because it is being dropped. The rest is right. Just ignore the negative sign when finding the square root because the negative just describes that the velocity is downwards (you can apply a negative afterward if you like). Also remember you can define downward as the positive direction and put g in as a positive number so that the velocity comes out positive.

Also remember to put units in. g is not equal to 9.8, it is equal to 9.8 m/s^2. Putting units on every value will help you to determine whether an equation makes sense.

Last edited: Oct 11, 2006
6. Oct 11, 2006

XPX1

xf^2=0+2(9.8)(18-0)

xf^2=352.8

square root of 352.8 is 18.78297101 or 18.8

so that would be the answer correct?

Also, is velocity the same as speed?

And, 18 had nothing to do with velocity, because it was not say it was moving down at 18m/s?

18 only related to the x final position because wherever it was, it has to go 18 meters right?

Thanks ALOT for your help BishopUser

7. Oct 11, 2006

BishopUser

you changed the equation you are calculating vf not xf, but yes the final velocity is 18.8 m/s. Velocity is speed in a given direction (velocity can be negative but speed cannot be). In this case you set the zero level at the bird and consider the initial position to be 0 and the final position to be 18 m since that's how far it has to falls.

So the final answer would be 18.8 m/s downward. You can signify that it is downward by defining a coordinate system in which downward is either positive or negative. Chances are your teacher would expect you to have -18.8 m/s since most students automatically assume that there is a predefined coordinate system (where downward is negative) when they should be creating their own

Last edited: Oct 11, 2006