Rock free fall

  • Thread starter Neophyte
  • Start date
  • #1
44
0

Homework Statement


A stone is thrown upward from ground level. The initial speed is 32 ft/sec
a) In how many seconds will the stone hit the ground?
b) How high will it go?
c) With what minimum speed should the stone be thrown so as to reach a height of 36 ft?


Homework Equations



y(t) = -16(t)2 + v0(t) + y0

The Attempt at a Solution



A) y(t) = -16(t)2 + 32(t) +0

0 = -16(t)2 +32
t = 2 sec till it hits the ground

B)y'(t) = v(t) = -32t + 32
0 = -32t +32
t = 1 sec to reach max

-16 +32 = 16 ft max

C) y(t) = 36 ft y(t) = -16(t)2 +32(t)
-16(t)2 +32(t) = 36
+16(t)2 -32(t) +36 = 0
4(4(t)2 - 8(t) + 9 = 0


Then I assumed I was doing it wrong : /

I continued onward to get (t-1)2 = -(5/4)
t = 41/16
v(41/16) = -32(41/16) +32 = -50 = 50 ft/s

I got part c wrong naturally, any help on where it went wrong would be appreciated. Or if I did the first two wrong and just got the right answer. Either way, thanks for your time.
 
Last edited:

Answers and Replies

  • #2
Cyosis
Homework Helper
1,495
0
This belongs in the introductory physics forums.

You did A and B correctly.

C is a bit sloppy although not necessarily the wrong approach. You calculated t incorrectly, and when you're using [itex]v=v_0-gt[/itex], you use [itex]v_0=32 \;\;ft/sec[/itex], but you're trying to solve [itex]v_0[/itex]! [itex]v_0=32 \;\;ft/sec[/itex] belongs to the previous question and of course if the initial velocity is the same as in part a then the maximum height will be the same so this cannot be true.

The best way would be to start with the formula for the velocity, so you don't have to use the ABC formula.

[itex]v=v_0-gt[/itex], what value is v at its maximum height?
 
  • #3
44
0
[itex]v=v_0-gt[/itex], what value is v at its maximum height?
Velocity = 0 at max so the initial velocity is equivalent to gravity(t)?

0 = V0 - 16t

16t = V0

or V0/16 = t
:uhh:
 
Last edited:
  • #4
Cyosis
Homework Helper
1,495
0
Since you have t now you can plug it into y(t) and solve for v0.
 
  • #5
44
0
So

y(V0/16) = -16(V0/16)2 + V0(t)

y(V0/16) = (V0)2/16 + V0(V0/16)

y(V0/16) = (V0)2/16 + (V0)2/16
y(V0/16) = 2V02/16

36 = 2V02/16

576 = 2V02

288 = V02

16.9 = V0

:cry: There is no hope it seems.

Step X) Answer is 48 b/c it is in the back of book, thus I win for now.
 
  • #6
Cyosis
Homework Helper
1,495
0
Velocity = 0 at max so the initial velocity is equivalent to gravity(t)?

0 = V0 - 16t

16t = V0

or V0/16 = t
Since I don't use the imperial system I missed this at first, but isn't g 32 ft/sec in the imperial system?

y(V0/16) = -16(V0/16)2 + V0(t)

y(V0/16) = (V0)2/16 + V0(V0/16)
Where did the minus go?
 
  • #7
44
0
g is 32 ft /s but the formula is y(t) = - (1/2)gt2 + V0 Y0

The minus did vanish but then wouldnt I get y(t) = 0? Well 36 = 0

-V02/16 + V02/16
 
  • #8
Cyosis
Homework Helper
1,495
0
g is 32 ft /s but the formula is y(t) = - (1/2)gt2 + V0 Y0

The minus did vanish but then wouldnt I get y(t) = 0?
Yes but you didn't use that formula, you used

[itex] v=v_0-gt[/itex] , what value is v at its maximum height?
to determine t.
 
  • #9
44
0
Finally, I think I love you tbh.

V0/32 = t

-16V02/1024 + V02/32

-16V02/1024 + 32V02/1024

36 = 16V02

36864 = 16V0

2304 = V0

48 ft/s = V0

Thank you for your help, I really appreciate it as I was close to burning the book.:blushing:
 
  • #10
Cyosis
Homework Helper
1,495
0
You're welcome but you should really be less sloppy. Your answer is correct but if you wrote it down like that an exam I fear you would get very little points.

V0/32 = t

-16V02/1024 + V02/32

-16V02/1024 + 32V02/1024
All good so far.

then

36 = 16V0^2
You multiplied the right side by 1024, but didn't multiply the left side.

It should be [itex]36*1024=16v_0^2[/itex]

36864 = 16V0
Here the 1024 is back so the left hand side is correct now, but now you lose a square on the right hand side.

It should be

[itex]36864=16v_0^2[/itex]

2304 = V0
You divided by 16 on both sides correctly, but again the square is gone.

It should be [itex]2304=v_0^2[/itex]

48 ft/s = V0
I know you took the square root on both sides here, but as you've written the last two equations down it reads [itex]v_0=2304ft/s=48ft/s[/itex] which obviously isn't right.

Tip: I personally prefer to not fill in the numbers until the end. This way you're less prone to make mistakes. We know that [itex]t=\frac{v_0}{g}[/itex], therefore [itex]y_{max}=\frac{v_0^2}{g}-\frac{1}{2}\frac{v_0^2}{g}=\frac{1}{2}\frac{v_0^2}{g}[/itex]. Solve for v_0 gives [itex]\sqrt{2 g y_{max}}=v_0[/itex]. It is exactly the same method but a lot cleaner. Now enter the numbers and obtain your answer.
 
Last edited:

Related Threads on Rock free fall

Replies
1
Views
4K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
17K
  • Last Post
Replies
6
Views
1K
Replies
3
Views
714
  • Last Post
Replies
1
Views
6K
Replies
2
Views
2K
  • Last Post
Replies
7
Views
1K
Top