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Very simple position, velocity, acceleration question

  • #1
264
0

Homework Statement



A stone is dropped from the top of a 400 m tower. (Acceleration due to gravity is -9.8 m/s2. Ignore air resistance. Give your answers correct to two decimal places.)

Homework Equations





The Attempt at a Solution



so part a asks for an equation to model the position of the ball which is:

[tex] h(x)= \frac {-9.8t^{2}}{2}+400 [/tex]

which is correct, the next question asks how long does it take to hit the ground:


[tex] h(x)= \frac {-9.8t^{2}}{2}+400=0 [/tex]

[tex] \frac {-9.8t^{2}}{2}=-400 [/tex]

[tex] -9.8t^{2}=-800 [/tex]

[tex] t^{2}= \frac {4000}{49} [/tex]

[tex] t= \sqrt{ \frac{4000}{49}} \approx 9.04s [/tex]

9.04s is also correct, so the next question asks at what speed does the ball hit the ground, so i take the derivative of h(x) to find this, correct?

[tex] h'(x) = -9.8t [/tex]

[tex] h'(x) = -9.8(9.04) \approx -88.592 \approx -88.6 [/tex]

but i have tried -88.6, -88.59, 88.6, and 88.59 so did I do something wrong here? I must be missing something...

thanks!
 

Answers and Replies

  • #2
Hmm, I don't see a problem with your method. Perhaps the (I'm assuming this is mastering physics or some other software oriented assignment) program wants you to use:

[tex]v_{f}^{2}=v_{i}^{2}+2*g*d[/tex]

That gives me a final velocity of 88.54. Though I would personally use the method you are trying.
 
  • #3
264
0
This is actually just a calculus class, i got the first equation easily because i knew

[tex] x = \frac {1}{2}a t^{2} + v_{0}t + x_{0} [/tex]

from my physics class. So I shouldn't be expected to use any other formula than what it asked me to derive from the question itself. Perhaps the software is wrong then? lol I will ask my teacher tomorrow then, thanks for confirming what I suspected from the start.

edit: we use an online homework app for some sections called "Web Assign", if you have ever heard of it. Not that it matters but I was clarifying that I am indeed putting the answers into the computer.
 
  • #4
eumyang
Homework Helper
1,347
10
I think the problem is that you rounded too soon. When you found the time it took for the stone to hit the ground:
[tex]t= \sqrt{ \frac{4000}{49}} \approx 9.04s[/tex]
Don't round here before plugging this value into the h'(t) equation (shouldn't it be h(t) and not h(x)?). Use as many digits as you can. Windows Calculator give me for t:
t ≈ 9.0350790290525123771396958412363
So if you plug this in the h'(t) equation you get
[tex]\begin{aligned}
h'(9.0350790290525123771396958412363) &\approx -9.8 \cdot 9.0350790290525123771396958412363 \\
& \approx -88.543774484714621295969019244116 \; m/s
\end{aligned}[/tex]
So -88.54 was probably the answer that WebAssign wanted.
 

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