# Thread: Gravity motion mechanics 1

1. ## 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

2. 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

3. 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.

4. Hello, gracey!

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

5. 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...

6. 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

7. 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?

8. 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