Very simple position, velocity, acceleration question

[Asphyxiated]

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:

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

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


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

\frac {-9.8t^{2}}{2}=-400

-9.8t^{2}=-800

t^{2}= \frac {4000}{49}

t= \sqrt{ \frac{4000}{49}} \approx 9.04s

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?

h'(x) = -9.8t

h'(x) = -9.8(9.04) \approx -88.592 \approx -88.6

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!

 

[CalcYouLater]

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:

v_{f}^{2}=v_{i}^{2}+2*g*d

That gives me a final velocity of 88.54. Though I would personally use the method you are trying.

 

[Asphyxiated]

This is actually just a calculus class, i got the first equation easily because i knew

x = \frac {1}{2}a t^{2} + v_{0}t + x_{0}

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.

 

[eumyang]

I think the problem is that you rounded too soon. When you found the time it took for the stone to hit the ground:
t= \sqrt{ \frac{4000}{49}} \approx 9.04s
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
\begin{aligned}<br /> h&#039;(9.0350790290525123771396958412363) &amp;\approx -9.8 \cdot 9.0350790290525123771396958412363 \\<br /> &amp; \approx -88.543774484714621295969019244116 \; m/s<br /> \end{aligned}
So -88.54 was probably the answer that WebAssign wanted.