# A rock is dropped from the top floor of a tower at a height of 420m?

## How do you find the distance of the rock from the ground at any time (t)? How do you figure out how long it takes for the rock to reach the ground and what will its velocity be?

Nov 21, 2017

Part 1
$s \left(t\right) = 420 - \frac{1}{2} \text{g} {t}^{2}$

Part 2a
$t = 9.26 s$

Part 2b
$v = - 90.7 \text{ m/} s$

#### Explanation:

The equation used for position $\left(s\right)$ as a function of time is:

$s \left(t\right) = {s}_{0} + {v}_{o} t + \frac{1}{2} a {t}^{2}$

where ${s}_{0}$ is initial position, ${v}_{0}$ is initial velocity, and $a$ is (constant) acceleration

If you've done physics before, there's a very good chance you've seen this equation. Either way, having this memorized will help you solve problems like this.

Part 1
Since you're dropping the rock (i.e. not throwing it), the rock has no initial velocity. Hence, we can just make that zero, leaving:

$s \left(t\right) = {s}_{0} + \frac{1}{2} a {t}^{2}$

Also, your only source of acceleration is due to gravity (g = 9.8 m/${s}^{2}$), so you can further simplify this to:

$s \left(t\right) = {s}_{0} - \frac{1}{2} \text{g} {t}^{2}$

The negative sign simply takes into account that the rock's acceleration vector is pointing in the negative direction (down), since we typically define up as positive.

The equation represents position as a function of time, so all you have to do is plug in your specific ${t}_{f}$:

color(blue)(s(t_f) = $\textcolor{b l u e}{420 \text{ m" - 1/2(9.8 " m/} {s}^{2}} {\left({t}_{f}\right)}^{2}$

And that will give you your position at any time ${t}_{f}$.

Part 2a
When the rock hits the ground, we know that the final position will be $0$, since that's how we defined our coordinate system. Therefore:

$s \left(t\right) = 0 =$ 420 m $- \frac{1}{2} \left(9.8 \text{ m/} {s}^{2}\right) {\left(t\right)}^{2}$

Now it's just a matter of solving for $t$:

$420 - \frac{1}{2} \left(9.8\right) {\left(t\right)}^{2} = 0$

$\frac{1}{2} \left(9.8\right) \left({t}^{2}\right) = 420$

$\implies \textcolor{b l u e}{t = \sqrt{\frac{840}{9.8}} = 9.26 s}$
*rounding to 3 significant digits.

Part 2b
So now we have the time at which it hits the ground. What about the velocity when it hits the ground? Well, we can recall a very handy fact about the relationship between position and velocity, which is that:

$s ' \left(t\right) = v \left(t\right)$.

i.e. velocity is the first derivative of position. So, we can actually use our position function to derive velocity as a function of time (this is just a basic power rule computation):

$v \left(t\right) = s ' \left(t\right) = - \text{g} t$

Since we know the time at which the ball hits the ground, we can just plug that in and find the velocity:

$v \left(9.26\right) = - 9.8 \left(9.26\right)$

$\implies \textcolor{b l u e}{v = - 90.7 \text{ m/} s}$
*rounding to 3 significant digits

When performing that final step, make sure you use your unrouded value (i.e. 9.2582....$s$) for time it takes to fall. This ensures you have a much greater degree of accuracy.

Hope that helped :)

Nov 21, 2017

See below.

#### Explanation:

The general equations are:

$s = u t + \frac{1}{2} a {t}^{2} \textcolor{w h i t e}{88} \left[1\right]$

$v = u + a t \textcolor{w h i t e}{88} \left[2\right]$

Where $s =$distance ,$u =$ initial velocity , $v =$ final velocity , $t =$ time and $a =$ acceleration.

For the problem given, you would use [ 1 ] for the first part.

When dropping an object from the tower the initial velocity would be 0. The value of $a$ will be $g$. this is known as the acceleration due to gravity and is approximately $9.8 m {s}^{- 2}$.

So the distance the object falls in time $t$ will be given by:

$s = \left(0\right) t + \frac{1}{2} \left(9.8\right) {t}^{2}$

For the time it takes the rock to hit the ground:

Distance travelled is $420 m$

So using [ 1 ]:

$420 = \left(0\right) t + \frac{1}{2} \left(9.8\right) {t}^{2}$

$\frac{1}{2} \left(9.8\right) {t}^{2} = 420 \implies {t}^{2} = \frac{840}{9.8} \implies t = \sqrt{\frac{840}{9.8}} = \textcolor{b l u e}{9.26 s}$

( 2 .d.p.)

Now we know the time it take to reach the ground, we can use [ 2 ] to find the final velocity

$v = \left(0\right) + \left(9.8\right) \left(9.26\right) = \textcolor{b l u e}{90.75 \textcolor{w h i t e}{88} m {s}^{- 2}}$

There are several other equations for these types of problems and rearrangements of the ones above. You will find them in text books or online.