Question

When a volcano erupts, it emits smoke and lava in all direction with a velocity of 280m/s. The height of volcano is 5 km amd there is a house located at a distance of 8 km from the centre of the mountain. 5s after an eruption, a group of people start fleeingfrom the house in a car that can run at a maximum speed of 90km/h.

Assuming that they travel in a straight line away from the volcano, what additional distance should they have covered to escape the fireball?

Answer 1

When a volcano erupts, it emits smoke and lava in all direction with a velocity of $280\mathrm{m / s}$ . The height of volcano is $5\mathrm{km}$ and there is a house located at a distance of $8\mathrm{km}$ from the centre of the mountain. 5s after an eruption, a group of people start fleeing from the house in a car that can run at a maximum speed of $90\mathrm{km / h}$ .

Assuming that they travel in a straight line away from the volcano, what additional distance should they have covered to escape the fireball?

Solution:

$h = 5km = 5000m$ - height of volcano;

$T = 5s$ time after eruption, a group of people start fleeing from the house in a car;

$V = 280\frac{m}{s} - \text{velocity of the lava};$

$U = 90\frac{km}{h} = 25\frac{m}{s} -$ maximum speed of the car;

$L = 8km = 8000m - a$ distance from the centre of the mountain;

$d$ - additional distance should group have covered to escape the fireball.

We need to consider the fireball, which emitted from volcano at an angle of 45 degrees, because at this angle you can find the maximum distance that the lava can reach:

Equation of motion of fireball: $(\alpha = 45{}^{\circ})$

$y:0 = h + V_{y}t - \frac{gt^{2}}{2};$ where $t -$ time before falling to the ground

$x: S_{1} = V_{x}t$

$V_{y} = V\sin \alpha ;V_{x} = V\cos \alpha \rightarrow$

$$h = \frac {g t ^ {2}}{2} - V t \sin \alpha \quad (1)$$ $$S _ {1} = V t \cos \alpha$$ $$t = \frac {S _ {1}}{V \cos \alpha} \quad (2)$$ $$(2) \text{in} \quad (1)$$ $$h = \frac {g \left(\frac {S _ {1}}{V \cos \alpha}\right) ^ {2}}{2} - \frac {S _ {1}}{\cos \alpha} \sin \alpha$$ $$h = \frac {g S _ {1} ^ {2}}{2 V ^ {2} \cos^ {2} \alpha} - S _ {1} \tan \alpha$$

Substitute numerical values to find the maximum distance $S_{1}$: $(\cos \alpha = \sin \alpha = \frac{1}{\sqrt{2}}; \tan \alpha = 1)$

$$5000 = \frac {9.8 S _ {1} ^ {2}}{\left(280 \frac {m}{s}\right) ^ {2}} - S _ {1}$$ $$0.000125 \cdot S _ {1} ^ {2} - S _ {1} - 5000 = 0$$

We have a quadratic equation whose roots are:

$$S _ {1} = -3483.31; 11483.3$$ $$S _ {1} > 0, \text{distance can not be negative} \rightarrow S _ {1} = 11483.3m$$ $$t = \frac {S _ {1}}{V \cos \alpha} = \frac {11483.3m}{280 \frac {m}{s} \cdot \frac {1}{\sqrt {2}}} = 58s$$

Now we can consider the motion of the car. Distance that the car has passed:

$$S _ {2} = U \cdot (t - T) = 25 \frac {m}{s} (t - 5s) = 25 \frac {m}{s} (58s - 5s) = 1325m$$

With $S_{1}$ and $S_{2}$ we can find additional distance should group have covered to escape the fireball:

$$d = S _ {1} - L - S _ {2}$$ $$d = 11483.3m - 8000m - 1325m = 2158m$$

Answer: additional distance should group have covered to escape the fireball: $d = 2158m$.