Question

A rock is dropped from rest from the roof of a building 100 m high. What is the velocity of the rock at the moment it reaches the ground? How long does it take to hit the ground?

Answer 1

The final velocity of the rock is `-44.3m/s`, and its fall time is `4.52s`.

Explanation:

There are at least two ways to solve this, one being with kinematics and the other with a combination of kinematics and energy conservation. These are just the first two that come to mind. I'll give an explanation of both methods.

Method 1: Kinematics

This is a projectile motion problem which can be solved using kinematics. Because the rock is dropped from rest, we know its initial velocity is `0` (`v_i=0`). We are also given that it is dropped from a height of `100m` (`y_i=100m`). Because the object hits the ground, we can define its final height as `0` (`y_f=0`). Finally, because the object experiences free-fall, we know that its acceleration is equal to `-g`, or `-9.8m/s^2`.

We can use this kinematic equation to solve for the final velocity, `v_f`:

`v_f^2=v_i^2+2a_yΔy`

where `Δy=y_f-y_i`. Using `v_i=0` as determined above, solving for `v_f` gives:

`v_f=sqrt(2a_yΔy)`

Using our known values:

`v_f=sqrt(2(-9.8m/s^2)(0m-100m))`

`v_f=44.3m/s` downward or `-44.4m/s`.

To find the fall time, we can use this kinematic equation:

`y_f=y_i+v_(iy)Δt+1/2a_yΔt^2`

Because `y_f` and `v_(iy)` are both 0, we can rearrange to solve for Δt:

`Δt=sqrt((-2y_i)/a_y)`

`Δt=sqrt((-2(100m))/(-9.8m/s^2)`

`Δt=4.52s`

You could also use `v_f=v_i+a_yΔt` after finding the final velocity.

Method 2: Energy Conservation/Kinematics

`U_(gi)+K_i=U_(gf)+K_f`

Where `U_g` is the initial gravitational potential energy (initial and final), and `K` is the kinetic energy (initial and final).

Kinetic energy is given by `K=1/2mv^2` and gravitational potential energy is given by `U_g=mgh`.

As the rock is not moving initially (i.e. at rest), for our intents and purposes it possesses only gravitational potential energy. When it is dropped, that gravitational potential energy is transformed into kinetic potential energy as it falls. Just before the rock hits the ground, it has only kinetic energy (`h≈0`). Therefore, our equation becomes:

`U_(gi)=K_f`

`mgh_i=1/2mv_f^2`

We can see that mass cancels, as it is present on both sides, giving:

`gh_i=1/2v_f^2`

Solving for `v_f`,

`v_f=sqrt(2gh_i)`

`v_f=sqrt(2(9.8m/s^2)(100m))`

`v_f=44.3m/s`

You would then take that final velocity and use a kinematic equation to find Δt, either the one used above in the kinematics method or `v_f=v_i+a_yΔt`.

Hope that helps!

Answer 2

The final velocity is `"-44.3 m/s"`.

The rock will take `"4.52 s"` to fall.

Explanation:

To find the final velocity, use the kinematic equation `v_f^2=v_i^2+2ad`.

`v_i=0`
`a="-9.8 m/s^2"`
`d="-100 m"`
`v_f="???"`

Substitute the given values into the equation.

`v_f^2=0+2*-9.8 m/s^2*(-100"m")`

`v_f^2=1960"m"^2/(s^2")`

`v_f=+-sqrt((1960"m"^2/s^2))="-44.3 m/s"` (final velocity is in a negative direction)

To determine the length of time it takes for the rock to fall, use the kinematic equation `v_f=v_i+at`. Since `v_i=0`, we can leave it out of the equation. `v_f=at"`

Rearrange the equation to isolate `t`.

`t=v_f/a`

`t=(-44.3cancel"m"/"s")/(-9.8cancel"m"/(cancel(s"^(2))` `=` `4.52 "s"`