Falling objects - finding the speed

Question

I am trying to work out how fast water will be falling by the time the water hits the ground.

If it starts 100m high how fast would it be travelling and why?

With the acceleration because of gravity to be 9.8

What would the equation be?

I think it has something to do with $speed^2 = height*acceleration(9.8)$.

Answer 1

You should have been taught the SUVAT equations of motion.

If the water is stationary then $u=0 \, \mathrm{m/s}$, the acceleration due to gravity is $-9.8 \,\mathrm{m/s^2}$. The displacement of the water is $-100 \, \mathrm m$, and you want to know the final velocity, i.e. you want $v$.

HINT: Use the equation $v^2 = u^2 + 2as$.

Answer 2

The rate of acceleration is:

$a = 9.8\,m/s^2$

which means the speed at which an object is falling is:

$v(t) = t*9.8\,m/s$

(so, for example, after three seconds the water will be falling at a rate of $3 *9.8\,m/s = 29.4 \text{ meters/second}$).

to find the formula for the distance an object has fallen we take the anti-derivative to get:

$D(t) = t^2/2*9.8\,m = t^2*4.9\,m$.

(so, for example, after three seconds the water will be falling at a rate of $29.4 \text{ meters/second}$, and it would have fallen $3^2*4.9\text{ meters} = 44.1 \text{ meters}.$)

So if $D(w) = w^2*4.9\text{ meters} = 100 \text{ meters}$, what is $w$?

Then to find out how fast it is falling, what is $v(w) = w*9.8\text{ m/sec}$?

Answer 3

$$\text{h}=\frac{\text{gt}^2}{2}\Longleftrightarrow2\text{h}=\text{gt}^2\Longleftrightarrow\frac{2\text{h}}{\text{g}}=\text{t}^2\Longleftrightarrow t=\sqrt{\frac{2\text{h}}{\text{g}}}$$

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$$\text{v}=\text{gt}\Longleftrightarrow\text{v}=\text{g}\cdot\sqrt{\frac{2\text{h}}{\text{g}}}\Longleftrightarrow\text{v}=\sqrt{\text{g}^2}\cdot\sqrt{\frac{2\text{h}}{\text{g}}}\Longleftrightarrow\text{v}=\sqrt{\frac{2\text{g}^2\text{h}}{\text{g}}}\Longleftrightarrow$$ $$\text{v}=\sqrt{\frac{2\text{g}\text{h}}{1}}\Longleftrightarrow\text{v}=\sqrt{2\text{g}\text{h}}$$

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So, when the fall distance is $100$ meter:

• $$\text{v}=\sqrt{2\text{g}\cdot100}=\sqrt{200\text{g}}=10\sqrt{2\text{g}}$$
• $$t=\sqrt{\frac{2\cdot100}{\text{g}}}=\sqrt{\frac{200}{\text{g}}}=\frac{10\sqrt{2}}{\sqrt{\text{g}}}$$

Assuming the gravitational acceleration as $\approx 9.81$:

• $$\text{v}\approx10\sqrt{2\cdot9.81}\approx44\space\text{m/s}$$
• $$t\approx\frac{10\sqrt{2}}{\sqrt{9.81}}\approx4.5 s$$