Question 1

Final Velocity ($v$)

Accepted Answer

is the speed of the object at any specific point in time during its fall, most commonly calculated at the exact instant before it impacts the ground. It is crucial to understand that in physics, the final velocity is never zero just because the object hits the floor and stops; final velocity refers to the maximum speed achieved immediately prior to the collision. Displacement or Height ($h$ or $y$) refers to the total vertical distance the object travels during the fall, typically measured in meters. Unlike distance, which is a scalar quantity, displacement is a vector, meaning it has a direction. In many free fall calculations, we define the starting point as zero and the downward direction as positive, making the height fallen a positive number.

Question 2

Velocity as a function of time:

Accepted Answer

$v = v_0 + gt$
This equation tells us how fast the object is moving after a specific amount of time has passed. If you drop an object from rest ($v_0 = 0$), the formula simplifies to just $v = gt$.

Question 3

Position as a function of time:

Accepted Answer

$h = v_0t + \frac{1}{2}gt^2$
This equation determines how far the object has fallen after a certain number of seconds. Again, if the object is dropped from rest, the initial velocity term ($v_0t$) becomes zero, simplifying the equation to $h = \frac{1}{2}gt^2$. You can algebraically rearrange this simplified formula to solve for time if you know the height: $t = \sqrt{\frac{2h}{g}}$.

Question 4

Velocity as a function of position:

Accepted Answer

$v^2 = v_0^2 + 2gh$
This is the time-independent equation. It is incredibly useful when you know the drop height but do not know, and do not care to find, the time it took to fall. If dropped from rest, it simplifies to $v^2 = 2gh$, which means the final velocity is $v = \sqrt{2gh}$.

Question 5

Step 1: Calculate the Time to Impact

Accepted Answer

We will use the rearranged position equation: $t = \sqrt{\frac{2h}{g}}$.
First, multiply the height by 2: $2 	imes 125 	ext{ meters} = 250 	ext{ meters}$.
Next, divide that result by gravity: $250 / 9.81 = 25.484$.
Finally, take the square root of that number: $\sqrt{25.484} = 5.048 	ext{ seconds}$.
The rock will be in free fall for exactly 5.048 seconds.

Question 6

Step 2: Calculate the Final Velocity

Accepted Answer

We can use the time we just found in the first equation ($v = gt$), or we can use the time-independent equation ($v = \sqrt{2gh}$) to verify. Let us use both to prove the math is sound.
Using time: $v = 9.81 	ext{ m/s}^2 	imes 5.048 	ext{ s} = 49.52 	ext{ m/s}$.
Using height: $v = \sqrt{2 	imes 9.81 	imes 125} = \sqrt{2452.5} = 49.52 	ext{ m/s}$.
The rock will hit the ground traveling at $49.52 	ext{ meters per second}$ (which is approximately $110.7 	ext{ miles per hour}$).

Question 7

Does a heavier object fall faster than a lighter object?

Accepted Answer

In a perfect vacuum, absolutely not. Gravity accelerates all objects at the exact same rate ($9.81 	ext{ m/s}^2$) regardless of their mass. This is because an object with more mass requires more force to accelerate, and the Earth's gravity perfectly scales up its pulling force to match the increased mass, canceling out any difference. However, in the real world with air resistance, a heavier, denser object will often hit the ground first because it has enough momentum to overcome aerodynamic drag more effectively than a lighter object of the same size, such as a rock falling faster than a crumpled piece of paper.

Question 8

How fast is a human falling at terminal velocity?

Accepted Answer

For a human skydiver falling in a standard belly-to-earth position, terminal velocity is typically reached after about 12 seconds of free fall, which equates to a drop of roughly 450 meters (1,500 feet). At this point, the air resistance pushing up equals the gravity pulling down, and the skydiver stops accelerating. The standard terminal velocity for an average human in this posture is approximately $54 	ext{ meters per second}$, or $120 	ext{ miles per hour}$. If the skydiver tucks their arms and dives headfirst, reducing their aerodynamic profile, they can reach speeds exceeding $89 	ext{ m/s}$ (200 mph).

Question 9

What happens if you throw the object downward instead of just dropping it?

Accepted Answer

If you forcefully throw an object downward, you are altering its initial velocity ($v_0$). Instead of starting at $0 	ext{ m/s}$, it starts at whatever speed your arm imparted to it. For example, if you throw a ball downward at $10 	ext{ m/s}$, the kinematic equation for velocity becomes $v = 10 + 9.81t$. The object will still accelerate at the normal rate of gravity ($9.81 	ext{ m/s}^2$), but because it started with a "head start" in speed, it will cover the distance faster and hit the ground with a higher final velocity than an object simply dropped from rest from the same height.

Question 10

How do you calculate the depth of a well using a dropped stone and the speed of sound?

Accepted Answer

This is a classic, complex physics problem. When you drop a stone down a dark well and wait for the splash, the total time you measure consists of two separate events: the time it takes the stone to fall down, and the time it takes the sound to travel back up. To solve this, you must set up a system of equations. The time for the stone to fall is $t_1 = \sqrt{\frac{2h}{9.81}}$. The time for the sound to travel up is $t_2 = \frac{h}{343}$ (where 343 m/s is the speed of sound). Since your total measured time is $T = t_1 + t_2$, you substitute the equations in and solve the resulting quadratic equation for $h$. It requires advanced algebra, but it yields an incredibly precise measurement of the well's depth.

Question 11

Can an object or person survive hitting the ground at terminal velocity?

Accepted Answer

Under normal circumstances, impacting a solid surface at terminal velocity (120 mph) is instantly fatal due to massive blunt force trauma and the sudden deceleration destroying internal organs. However, there are exceedingly rare, documented edge cases of survival. In 1972, flight attendant Vesna Vulović survived a fall from 10,160 meters (33,330 feet) after her plane exploded. Her survival was attributed to being pinned inside a section of the fuselage that hit a heavily forested, snow-covered mountain at a shallow angle. The crushing of the fuselage, the snapping of the pine trees, and the deep snow acted as a series of crumple zones, artificially extending the duration of the impact and reducing the peak deceleration forces to barely survivable levels.

Question 12

Does gravity change depending on where you are on Earth?

Accepted Answer

Yes, standard gravity ($9.80665 	ext{ m/s}^2$) is an average, but local gravity varies slightly. Because the Earth spins, centrifugal force pushes outward strongest at the equator, slightly counteracting gravity. Additionally, the Earth bulges at the equator, meaning you are further from the center of the Earth's mass there than you are at the poles. Therefore, gravity is weakest at the equator ($9.780 	ext{ m/s}^2$) and strongest at the North and South Poles ($9.832 	ext{ m/s}^2$). Furthermore, gravity decreases slightly as your altitude increases, meaning an object dropped in high-altitude Denver will fall infinitesimally slower than an object dropped at sea level in Miami.

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