下落物体（一维）

Falling Objects (1D Motion)

When you drop an object, its initial velocity (Vo) is zero and it accelerates at $-9.8m/s^2$ as it is the acceleration due to gravity.

We use these equation while solving problems that involve 1D motion: $$V = V_o + at$$ $$X = V_ot + \frac{1}{2}a{t^2}$$

$V$ = Final Velocity $(m/s)$

$Vo$ = Initial Velocity $(m/s)$

$a$ = acceleration ${(m/s^2)}$

$t$ = time $(s)$

$X$ = displacement $(m)$

Example:

You drop a rock from the top of a $500m$ tall building. How long does it take to reach the ground?

What we know:

$$Vo = 0m/s$$ $$a = -9.8{m/s^2}$$

$$X = -500m$$ ($X$ is negative because the rock is displacing in the -y direction)

What we must find: $t$

Now we plug what we know into $X = Vot + \frac{1}{2}a{t^2}$ and solve for t

$$-500 = 0t + (\frac{1}{2})(-9.8){t^2}$$ $$-500 = (\frac{1}{2})(-9.8){t^2}$$ $$-1000 = (-9.8){t^2}$$ $$-102.041 = {t^2}$$ $$t\approx10.1015\text{seconds}$$

What is the velocity at impact (final velocity)?

We must plug in what we know into $V = Vo + at$ and solve for $V$

$$V = 0 - 9.8(10.1015)$$ $$V \approx -98.9949m/s$$

Test Yourself!

Suppose you have a ball in your hand. Suppose that you let go of the ball. Where would it go? It would fall down towards the Earth, thus classifying it as a "falling object." Let's look into this.

Let's create a universe to work in for ourselves. I'm going to set up the following parameters:

a) We are on Earth, which means that $g$ = 9.81 $m/s^2$

b) The ball is being held 20 meters off of the ground - pretend you're standing on a wall.

c) Friction and drag due to air are negligible, and there is no wind.

Now, we're ready to start!

You are standing on the wall, ball in hand, and, after making sure no one is standing under you, you release the ball. Since you do not apply any initial force, the initial velocity of the ball is simply 0 $m/s$.

However, there is a force acting on it - the force of gravity. Due to this, the ball's velocity increases as it falls. A formula that tells us what velocity the ball has after a certain period of time is $V=Vi+at$, where $V$ and $Vi$ are velocity and initial velocity, respectfully, $t$ is time, and $a$ is acceleration (in this case, it is equal to $g$).

So, let's figure out what the velocity of the ball is when it hits the ground. To do this, we need to know the time it takes for the ball to reach the ground, or in other words, travel 20 meters.

For this, we need to use another formula: $d=Vi*t+.5at^2$. $d$ stands for displacement in this formula, which is basically the shortest distance from the initial position to the final position of the ball. We can plug in the values that we have already to come up with this equation: $20=0*t+.5(9.81*t^2)$.

After doing some voodoo voodoo mathematics, we come to get a value of 2.019 seconds for $t$. If you rounded $9.81 m/s^2$ to $10 m/s^2$, this would come out to exactly 2 seconds, but rounding isn't the way I do things. Therefore, $t=2.019 s$.

Now, let's go back to finding the final velocity, $V$. Putting the new values into our equation, we get $V=0+(2.019*9.81)=19.81 m/s$.

Thus, the ball's final velocity, right as it hit the ground, is 19.81 $m/s^2$.

Now for a recap of the formulas:

a) $V=Vi+at$

b) $d=Vi*t+.5at^2$

An important thing to remember is that these formulas are based on the assumption that there is no wind, and that friction and drag are negligible.