Question

In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of impact records the fall of the object (see figure).

Answer 1

h=Vi*t+(1/2)a*t^2
Vi=0 (dropped object => initial velocity is 0)
h=(1/2)a*t^2

Answer 2

(a) Find the position function that yields the height of the object at time t assuming the object is released at time t = 0.
h(t) =



At what time will the object reach ground level? (Round your answer to three decimal places.)
t =

sec

(b) Find the rates of change of the angle of elevation of the camera when t = 1 and t = 2. (Round your answers to four decimal places.)
θ'(1) = rad/sec
θ'(2) = rad/sec

Answer 3

I got h(t) = 16t^2 +400 and t = 5.

Answer 4

a=g=-9.8 (the direction is downward)
h=(-9.8/2)t^2

Answer 5

now I'm trying to find the theata 1 and 2 :(.

Answer 6

tan() = h/500 = -16t^2 +400/500

Answer 7

Wait where did you get h(t) from?

Answer 8

I think your h(t) is wrong:
h=d=\[V_{i}*t+\frac{1}{2}a*t^2\]

Answer 9

na it's correct, I got it from what the practice question was explaining, but I cannot find the theatas...

Answer 10

It doesn't make much sense to me either lol :(.

Answer 11

Either your h(t) is very wrong, or I've never been taught proper physics or I didn't learn anything in physics.

Answer 12

@KingGeorge take a look.

Answer 13

Apparently this is calculus physics :P

Answer 14

here's the practice problem....

Answer 15

OHHHH

Answer 16

d=d_i+V_i*t+(1/2)at^2

Answer 17

I think his h(t) is correct. We're using feet not meters.

Answer 18

silly me >.<
Hate it when they do that.

Answer 19

As for the angles, I never could work with angles in physics.

Answer 20

hmm....

Answer 21

I think I got it lol :)

Answer 22

You think you did or you know you did?

Answer 23

nope I failed :(. LEts do this!

Answer 24

Ok, position is:
\[h_{f}=h_{i}+V_{i}*t+\frac{1}{2}a*t^2\]
The object is released from res, from 256 feet, so your equation becomes:
\[h_{f}=256+0+\frac{1}{2}a*t^2=\frac{1}{2}a*t^2+256\]

Answer 25

Acceleration due to gravity in feet is apparently about 32ft/s^2 so the final equation is:
\[h_f=-\frac{32}{2}*t^2+256\]
We're using -32ft/s^2 because acceleration is pointing downwards, thus it's negative.
Simplify final h and you get:
\[h_f=-16t^2+256\]

Answer 26

mhm mhm. I got all that, jus tkonfused with the theata 1-2.

Answer 27

Ok

Answer 28

I'll use u instead of theta (easier to type)
tan(u)=opposite over adjacent = h/500, right?

Answer 29

mhm

Answer 30

so u = arctan.

Answer 31

Yes, but keep it as tangent, it's simpler to deal with.

Answer 32

\[tan(u)=h/500=\frac{-16t^2+256}{500}\]
You want du/dt at t=1 and t=2

Answer 33

ok

Answer 34

\[tan(u)=\frac{16}{500}(-t^2+16)=\frac{4}{125}(16-t^2)\]
Implicit differentiation gives:
\[Sec^2(u)\frac{du}{dt}=\frac{4}{125}*(-2t)\]
Sorry typing it in matlab is a pain and not typing it in matlab is ugly

Answer 35

So:
\[\frac{du}{dt}=\frac{-8t}{125}*\frac{1}{Sec^2u}\]

Answer 36

Actually I think their way could have been easier, since now I ended up with a sec^2(u), but that's not a problem.
Sec^2-Tan^2=1, Sec^2=1+tan^2
Tan=(4/125)(16-t^2)
Sec^2=1+((4/125)(16-t^2))^2

Answer 37

\[\frac{du}{dt}=\frac{-8t}{125}*\frac{1}{(1+\frac{4}{125}(16-t^2))^2}\]

Answer 38

Now just plug in t=1, and t=2 and use a calculator to evaluate corresponding du/dts

Answer 39

Are you ok with it now?

Answer 40

so it's 16-? I thought it was 25- for 400. But I tried it that way and it's not right... it was 16- for 256 feet though..

Answer 41

−1000/( 15,625 + 16(25 − 1^2)^2)

Answer 42

Huh? You lost me, I got the same equation as them.

Answer 43

*as they did

Answer 44

*Always use proper English. Always, even in math.

Answer 45

I was loking through it apparently it was based on what the # was... IT was 16 for 256 feet. Then for 144 it was 9. 256/16 = 16 144/16 = 9.

Answer 46

I tried 400/16 and got 25.

Answer 47

−1000/( 15,625 + 16(16 − 1^2)^2) is what they had for 256 feet.

Answer 48

ohh shoot I messed up the square

Answer 49

\[\frac{du}{dt}=\frac{-8t}{125}*\frac{1}{(1+\frac{4}{125}(16-t^2))^2}\]
Should have been:
\[\frac{du}{dt}=\frac{-8t}{125}*\frac{1}{1+(\frac{4}{125}(16-t^2))^2}\]

Answer 50

t=1
http://www.wolframalpha.com/input/?i=%28%28-8*1%29%2F%28125%29%29*1%2F%281%2B%28%284%2F125%29%2816-1%5E2%29%29%5E2%29

Answer 51

t=2
http://www.wolframalpha.com/input/?i=%28%28-8*2%29%2F%28125%29%29*1%2F%281%2B%28%284%2F125%29%2816-2%5E2%29%29%5E2%29

Answer 52

Happy? lol

Answer 53

That's what I thought the original ones in the practice were, but i will try them lol.

Answer 54

not correct :(.

Answer 55

http://www.wolframalpha.com/input/?i=%28%28-8*2%29%2F%28125%29%29*1%2F%281%2B%28%284%2F125%29%2825-2%5E2%29%29%5E2%29

http://www.wolframalpha.com/input/?i=%28%28-8*1%29%2F%28125%29%29*1%2F%281%2B%28%284%2F125%29%2825-1%5E2%29%29%5E2%29

Answer 56

But that's what your practice test gave.

Answer 57

these were incorrect as well though :(.

Answer 58

the practice is correct by using 16, I think we are supposed to use 25, but it dind't work either so Idk :(.

Answer 59

I shouldn't have entered those #'s but w/e... 1 try left :(.

Answer 60

Why 25? Aren't we solving the practice question?

Answer 61

no we are solving the original....

Answer 62

That's what I've been saying lol.

Answer 63

omfg.... you mean i just wasted my time solving the stupid practice problem with wrong numbers?

Answer 64

Hang on, lol I need a break from this problem, I'll just eat dinner, and then write this out on paper and then take a picture and post it.

Answer 65

I'll bb in like 5 mins

Answer 66

Everything should be the same except the # inside the () I thought it was 25, but it's not liking that apparently...

Answer 67

Not everything is the same.. Ok, hang on I'm solving it on paper.

Answer 68

\[\frac{du}{dt}=\frac{-8t}{125}*\frac{1}{1+(\frac{4}{125}(25-t^2))^2}\]
ermm you're right, it was just 25

Answer 69

but it doesn't work WHAT THE FAWK!

Answer 70

Did at least du/dt of 1 work?

Answer 71

OMFG IT SAYS 4 DECIMAL PLACES NOT 3 OMFG LOL.

Answer 72

FACE SMASH!

Answer 73

Ok, I want to kill you...

Answer 74

I know :(.

Answer 75

<3

Answer 76

lol it's ok pal, I get that a lot with this crap... That's why I prefer real teachers (lol I'm lucky I've never had online math assignments), I run into this all the time when I help my friends with their webworks and what not.

Answer 77

Yeah it's a pain, I'm tired and I was wondering if I messed up. Everything was 3 places, and I thought it said 3 before. I Was freaking out like WTF!?

Answer 78

Thanks again :).

Answer 79

np