Gravity motion mechanics 1

• Sep 12th 2008, 10:29 AM
gracey
Gravity motion mechanics 1
Could do with some help

In a volcanic eruptioni a rock is ejected vertically from the crater and rises to a height of 25 kilometres. Find the speed with which it emerges from the crater, and time it takes to reach this height.

Would appreciate any help

thanks :)
• Sep 12th 2008, 10:52 AM
topsquark
Quote:

Originally Posted by gracey
Could do with some help

In a volcanic eruptioni a rock is ejected vertically from the crater and rises to a height of 25 kilometres. Find the speed with which it emerges from the crater, and time it takes to reach this height.

Would appreciate any help

thanks :)

There are four basic kinematics equations for motion under constant acceleration. You are going to use two of them here.

For the first part we know that
$\displaystyle v^2 = v_0^2 + 2a(y - y_0)$
So set the zero level at the top of the volcano and upward as positive. Notice that at the top of the motion the speed is 0 m/s. So we get
$\displaystyle 0 = v_0^2 - 2gy$
which you can solve for $\displaystyle v_0$.

For the second part we also know that
$\displaystyle v = v_0 + at$
and again v = 0 m/s so
$\displaystyle 0 = v_0 - gt$
You know $\displaystyle v_0$ from the first part, so now just solve for t.

-Dan
• Sep 12th 2008, 11:05 AM
arbolis
Just for fun, as $\displaystyle g$ (the gravitational force on the Earth's surface) is not constant over these $\displaystyle 25 \text{km}$, you can also find $\displaystyle v_0$ with the following formula : $\displaystyle E=\frac{1}{2}mv^2-\frac{GmM_E}{r}$. Considering the Earth as a sphere with radius$\displaystyle =6400 \text{km}$, $\displaystyle G=6.67\times10^{-11}\text{N}\frac{m^2}{kg^2}$ and $\displaystyle M_E=5.97\times 10^{24}\text{kg}$ I found that $\displaystyle v_0$ is worth $\displaystyle 695.84 \frac{m}{s}$. Compare this result with yours, I think mine should be a bit lesser than yours.
• Sep 12th 2008, 11:11 AM
Soroban
Hello, gracey!

Quote:

In a volcanic eruption, a rock is ejected vertically from the crater
and rises to a height of 25 kilometres.

Find the speed with which it emerges from the crater,
and time it takes to reach this height.

The equation is: .$\displaystyle h(t) \;=\;h_o + v_ot - 4.9t^2$

where: .$\displaystyle \begin{Bmatrix} h &=& \text{height of rock} \\ h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \end{Bmatrix}$ . measured in meters and seconds

Assume $\displaystyle h_o = 0$
. . We will measure the rock's height from the top of the volcano.

The equation is: .$\displaystyle h(t) \;=\;v_ot - 4.9t^2$

This is a down-opening parabola; its peak is at its vertex.
. . $\displaystyle v \:=\:\frac{\text{-}b}{2a}$

With $\displaystyle a = \text{-}4.9$ and $\displaystyle b = v_o$, we have: .$\displaystyle t \:=\:\frac{\text{-}v_o}{2(\text{-}4.9)} \:=\:\frac{v_o}{9.8}$ seconds. .[1]

The maximum height is: .$\displaystyle h\left(\frac{v_o}{9.8}\right) \;=\;v_o\left(\frac{v_o}{9.8}\right) - 4.9\left(\frac{v_o}{9.8}\right)^2 \;=\;\frac{v_o}{19.6}$ meters

Since the maximum height is 25 km = 25,000 m,
. . $\displaystyle \frac{v_o}{19.6} \:=\:25.000 \quad\Rightarrow\quad v_o \:=\:490,000\text{ m/sec}$

Therefore, the initial velocity is: .$\displaystyle \boxed{490\text{ km/sec}}$

Substitute into [1]: .$\displaystyle t \;=\;\frac{490,000}{9.8} \;=\;\boxed{50,000\text{ seconds}}$

• Sep 12th 2008, 11:17 AM
arbolis
What?! Soroban, an initial speed of almost $\displaystyle 500 \text{ km/s}$! And it takes $\displaystyle 50 sec$ to reach only $\displaystyle 25 \text{ km}$! You certainly mean $\displaystyle 490 \frac{m}{s}$ which surprises me because I thought it would have been more than $\displaystyle 496\frac{m}{s}$.
I don't see where is my error...
• Sep 12th 2008, 11:36 AM
topsquark
Quote:

Originally Posted by Soroban

The maximum height is: .$\displaystyle h\left(\frac{v_o}{9.8}\right) \;=\;v_o\left(\frac{v_o}{9.8}\right) - 4.9\left(\frac{v_o}{9.8}\right)^2 \;=\;\frac{v_o}{19.6}$ meters

Since the maximum height is 25 km = 25,000 m,
. . $\displaystyle \frac{v_o}{19.6} \:=\:25.000 \quad\Rightarrow\quad v_o \:=\:490,000\text{ m/sec}$

Therefore, the initial velocity is: .$\displaystyle \boxed{490\text{ km/sec}}$
There was a typo or something here.

Substitute into [1]: .$\displaystyle t \;=\;\frac{490,000}{9.8} \;=\;\boxed{50,000\text{ seconds}}$

$\displaystyle h\left(\frac{v_o}{9.8}\right) \;=\;v_o\left(\frac{v_o}{9.8}\right) - 4.9\left(\frac{v_o}{9.8}\right)^2 \;=\; \frac{v_0^2}{19.6}$

So
$\displaystyle \frac{v_0^2}{19.6} = 25000 \implies v_0 = 700~m/s$

-Dan
• Sep 12th 2008, 11:48 AM
arbolis
topsquark, I think you are wrong with Soroban, I graphed the functions of position versus time and my answer was closer to 25000 m while 522 m/s was closer to 13000 m. Can you check that out once again?
• Sep 12th 2008, 12:17 PM
topsquark
Quote:

Originally Posted by arbolis
topsquark, I think you are wrong with Soroban, I graphed the functions of position versus time and my answer was closer to 25000 m while 522 m/s was closer to 13000 m. Can you check that out once again?

You are correct. I must have punched the numbers in wrong. I have fixed it in my response.

The point is that Soroban forgot to square the $\displaystyle v_0$ in his height equation. And notice that 490 km/s is significantly higher than escape velocity, so it won't have a max. height. :)

-Dan