Question

Looking for the answer to this: What is the terminal velocity of an apple falling 50 feet from the tree, assuming standard earth's gravity?

Answer 1

Do you know what standard Earth gravity is in feet per second?

Answer 2

Also, are you considering any air resistance - it might have a slight effect at 50 feet?

Answer 3

32 feet per second per second.

Answer 4

No, air resisitance is assumed to be 0.

Answer 5

Great! You can do a two step process using the position equation and velocity equation separately, or use the combined equation so that time isn't a factor.

Do you have the relevant equations handy?

Answer 6

I know the position equation is s= 1/2 a * tsquared + vt + c, with c-0. Is the velocity eq. v-s/t?

Answer 7

I meant v=s/t.

Answer 8

v=at is for velocity. velocity = acceleration × time

Answer 9

If you sub the velocity equation into the position equation, you get a combined equation in the form \(\large v^2-v_0^2=2a(y-y_0)\)

Answer 10

[it might be worthwhile to try this substitution yourself to derive this formula - will be good algebra practice.]

Answer 11

This will be a pretty direct way to get final velocity since \(y_0 = 0\), and \(v=0.\)

Answer 12

So then, s = 1/2 a*t sdquared +at +c.....

Answer 13

If you like. It'll take longer that way, because you have to solve for t first, but it isn't hard and works just as well.

Answer 14

I assume your y is my s...

Answer 15

yes, s is pretty generic, I'm used to calling vertical displacement y and horizontal, x.

Answer 16

Still trying to arrive at your comb equation...

Answer 17

\(y=y_0+vt+0.5at^2\)

\(v=at \rightarrow t=v/a\)

\(\rightarrow y=y_0+v(v/a)+0.5a(v/a)^2\)

Answer 18

Sorry, that's a little ambiguous...

\(\large y_f=y_0+v_0t+0.5at^2\)

\(\large v_f=v_0+at \rightarrow t=(v_f-v_0)/a\)

\(\large \rightarrow y_f=y_0+v_0((v_f-v_0)/a)+0.5a((v_f-v_0)/a)^2\)

Answer 19

\(\large \rightarrow y_f-y_0=2(v_0v_f-v_0^2)/2a+(v_f^2-2v_0v_f+v_0^2)/2a\)

\(\large \rightarrow y_f-y_0=\frac{v_f^2-v_0^2}{2a}\)

Should be able to take it from there.

Answer 20

Other option is to solve \(0=50-16t^2\) for t, then put that into velocity equation, \(v=at\).

Answer 21

Thanks, this was very helpful....haven't had this stuff in 50 yrs.