## kgeorge1 3 years ago If you drop a turkey 1000 ft how long does it take to fall? Can you show the work?

1. kgeorge1

Ignore resistance

2. Lukecrayonz

7.9 seconds.

3. gezimbasha

The acceleration is 32.2ft/s^2. The equation is$t=\sqrt{\frac{2d}{g}}=\sqrt{\frac{2*1000}{32.2}} \approx 7.88$

4. Lukecrayonz

^

5. Lukecrayonz

OH, show the work. Right haha. I thought it was a joke question so I just put the answer.

7. gezimbasha

@cgreenwade2000 aww man.. we should've done that!

8. kgeorge1

I just got my test back and he said the answer is 5.57. That is what I originally thought and I only asked here to find out whether or not I did it right. So now I am really confused is the answer 7.88 or 5.57??

9. gezimbasha

It is 7.88 if you posted the right question. If you need proof of that and how to come to the answer from the basic laws of mechanics I can be happy to provide to you or your teacher. Just make sure you've made the right question here. If it is, I can provide mathematical derivations of how to get this answer starting from one simple assumption, that acceleration is constant and it is equal to 32.2.

10. gezimbasha

Also, if you could be as nice as we are and post an explanation of how did you conclude your original thought, namely that the answer is 5.57.

11. kgeorge1

$1000=32.2t ^{2}$

12. kgeorge1

That is what I did. and arrived at 5.57. That is the way my teacher taught it and I think he was wrong.

13. gezimbasha

Let us draw upon the fact that the horizontal and vertical components of a body that is falling are independent of each other. This is also known as Galileo's law of falling bodies. Now, let us assume that gravity is constant everywhere and a fall from a distance d that is solely dependent on the effect of gravity. Therefore we form the equation,$a= 32.2 = constant$Now, we know from mechanics that acceleration is the rate of change of velocity therefore we can put write this mathematically as such,$a = \frac{dv}{dt}$If we multiply by dt, and integrate the equation we get,$\int\limits_{0}^{t}adt=\int\limits_{v_0}^{v}dv$The result of this integration is this one, and this is one of the easiest integrals in mathematics.$v-v_0 = at$We put v0 on the RHS and we get,$v=v_0+at$Now we know that velocity is the rate of change of the distance, so we make this equation:$v = \frac{ds}{dt}$Let s=distance. Now multiply by dt, and input the v on the equation.$(v_0+at)dt=ds$Integrate,$\int\limits_{0}^{s}ds=\int\limits_{0}^{t}v_0dt+\int\limits_{0}^{t}atdt$This gives you,$s=v_0t+\frac{1}{2}at^2$We know that your case deals with zero initial velocity, so set v0=0. And we know that s=1000. But we want to solve for t(time).$s=\frac{1}{2}at^2$Multiply by 2 and divide by a.$t^2=\frac{2s}{a}$Take the square root and you get the correct equation.$t=\sqrt{\frac{2s}{a}}$Keeping in mind that a is constant and is equal to 32.2 and s is equal to 1000, we input hte numeric values and get.$t=\sqrt{\frac{2*1000}{32.2}}=\sqrt{\frac{2000}{32.2}}=\sqrt{62.1118}=7.8811$

14. kgeorge1

Awesome! I will have to print this out and take to my teacher :)

15. gezimbasha

Let us now compute the velocity that your turkey has when it hits the ground. Knowing that the y-component or the vertical component is subject to gravitational attraction of 32.2ft/s^2 and keeping in mind that gravity attracts downwards, we write the equation,$s_y=-\frac{1}{2}at^2$Take the derivative,$v_y=-at$Let's for the sake of the argument assume that t=7.8811 is the actual correct answer. We input that here and compute,$v_y=-32.2*7.8811=-253.77$Now, using the equation we've derived, namely $s=v_0t+\frac{1}{2}at^2$We create a scenario where we first throw the turkey 1000ft up, and then it falls down. To do this we know that the initial speed is 253.77[ft/s] so we input that as v0=253.77.$s(t)=253.77t-\frac{1}{2}32.2t^2$Note the minus sign in the equation comes from the fact that acceleration pulls downwards. Now let's plot this and see what happens. As you can see the object first goes to 1000ft from our throw, then goes down in a matter of 7.8811 seconds. Exactly equal to the time it takes to go to 1000ft, if we were to throw it.