# Finding speed of a rock, given height acceleration and velocity

• April 12th 2011, 10:00 AM
katyi
Finding speed of a rock, given height acceleration and velocity
I have no idea how to do this question, it was on a test so I dont know the correct answer and when the prof gave out the solutions for the test I was away, so any help with this would be greatly appreciated :)

A rock is thrown upward from the top of a 100m building at 10 m/s. At what speed will it hit the ground (the acceleration due to gravity is -9.8 m/s^2)

I know this is probably like the easiest question ever but I, having trouble seeing what to do.
• April 12th 2011, 10:10 AM
Carlow52
Are you familiar with the equation
V = u + at
If you google for this equation you will get some pointers to the answer

In passing why the minus in -9.8 m/s^2)

Its a 2 part question, you need to find out how far up the rock will go first
• April 12th 2011, 10:15 AM
katyi
The only formulas we were taught for these types of problems were:

s= -4.9 t^2 - V0t + S0

and then the derivative of that = velocity, so theyre similar to the one you stated. But Im confused because I have no idea what to plug in for t. :S
• April 12th 2011, 10:37 AM
Carlow52
Okay but I don't subscribe to 'plug and solve' maths, you need to understand the process, its not hard.

The equation you have provided looks like the second equation of motion which is better written

S = ut + 0.5att [ I don't have the super script to write it as t squared]

S = distance traveled,
u = initial velocity
a = acceleration
t = time

and for the first eq
V = velocity at time t under acceleration a with initial velocity of u

So for the first piece you know u, a and V, which is zero at top of flight.

So using the first equation you can get t and then you can calculate S from my second equation so for the next piece you have the total distance to ground...

Give this a shot
• April 12th 2011, 11:32 AM
katyi
Okay, I think I still did this wrong so bear with me (thank you very much by the way)

so i did:

v= u + at
10 = 100 + (-9.8) t
-90 = -9.8 t
-9.813 = t

then
s= ut + - 0.5 att

=100 (-9.813) + 0.5(-9.8)(-9.813)^2
= -981.3 + 413.204
= -568. 095

does this make sense? i guess since its traveling downward would speed be negative?
• April 12th 2011, 12:26 PM
Soroban
Hello, katyi!

Quote:

A rock is thrown upward from the top of a 100m building at 10 m/s.
At what speed will it hit the ground?
(The acceleration due to gravity is -9.8 m/s^2)

You know the formula: . $s \;=\;s_o + v_ot - 4.9t^2$

We are given: . $s_o = 100,\;v_o = 10$

. . So we have: . $s \;=\;100 + 10t - 4.9t^2$

"Hit the ground" means: $s = 0.$

. . Hence: . $100 + t0t - 4,9t^2 \:=\:0 \quad\Rightarrow\quad 4.9t^2 - 10t - 100 \:=\:0$

Quadratic Formula: . $t \;=\;\dfrac{\text{-}(\text{-}10) \pm \sqrt{(\text{-}10)^2 - 4(4.9)(\text{-}100)}}{2(4.9)}$

. . $\displaystyle t \;=\;\frac{10 \pm\sqrt{2060}}{9.8} \;=\;\begin{Bmatrix}5.65 \\ \text{-}3.61 \end{Bmatrix}$

The rock hits the ground 5.65 seconds after it was thrown.

The velocity of the rock is given by: . $v \:=\:10 - 9.8t$

When $t = 5.65,\; v \:=\:10-9.8(5.65) \:=\:-45.37$

The rock hits the ground at 45.37 m/s.

• April 12th 2011, 12:36 PM
Carlow52
No bother will trying to help here. The solution offered above seems to ignore the stone being thrown up but maybe not

You need to get a handle on what what here and maybe read both the question and my posts a bit closer.

I also would like you to pay attention to the units as we go

You write
v= u + at
10 = 100 + (-9.8) t

We are told that 10 is initial velocity and in the formula V is final velocity so .....:)

To get the show on the road

V = u + at

0 = 10 + (-9.81)t

which gives t = .. well u tell me

Lets say the answer was 1
Then

S = ut + 0.5att [ I don't have the super script to write it as t squared]

S = distance traveled,
u = initial velocity
a = acceleration
t = time

S = 10*1 + 0.5(-9.81)*1*1

=5m

So now the stone is 5 m above the top of the building, at zero velocity, and we know that

S = ut + 0.5att [ I don't have the super script to write it as t squared]

S = distance traveled,
u = initial velocity
a = acceleration
t = time

So solve for t
105 = 0t + 0.5*9.81tt

and this goes into
V = u + at
• April 12th 2011, 02:05 PM
topsquark
Quote:

Originally Posted by Carlow52
No bother will trying to help here. The solution offered above seems to ignore the stone being thrown up but maybe not

Soroban's solution is correct in all aspects. However I will note that there is a more straightforward way to obtain the solution.
$v^2 = v_0^2 + 2a(s - s_0)$

$v^2 = (10)^2 - 2g(0 - 100)$

and note that when you take the square root at the end you select the negative solution since we are assuming positive is upward.

-Dan
• April 12th 2011, 02:20 PM
Carlow52
In this case then whats the solution if the stone is just dropped from 100m?

Given the difficulty the OP was having with inserting the right values into simple equations a more step by step approach seemed a little more appropriate.
• April 12th 2011, 05:30 PM
Nicolasa77
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• April 12th 2011, 06:08 PM
topsquark
Quote:

Originally Posted by Carlow52
In this case then whats the solution if the stone is just dropped from 100m?

Given the difficulty the OP was having with inserting the right values into simple equations a more step by step approach seemed a little more appropriate.

Then, using my equation, v0 = 0 m/s.

I was simply pointing out that we don't need to find the max height to do the problem. It can be done in one step.

-Dan
• April 15th 2011, 12:34 PM
HallsofIvy
Essentially, topsquark is using "conservation" of energy. The potential energy, 100 meters above the ground, is 100mg where m is the mass of the rock and g= 9.8 m/s^2, the acceleration due to gravity. The kinetic energy, at 10 m/s, is (1/2)m(100)= 50m. The total energy is 100mg+ 50m= (980+ 50)m= 1030m Joules At the ground the potential energy is 0 so all of that energy have converted to kinetic energy: (1/2)mv^2= 1030 so v^2= 2060.