# How Do You Derive the Equation for a Vertically Thrown Stone from a Cliff?

• MHB
• karush
In summary, a stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high. The equation for the parabola representing its displacement is f(t) = -0.5(9.8)t^2+15.5t+75. The highest point will be reached after approximately 1.582 seconds. The model only refers to vertical displacement and the horizontal axis represents time. The t-axis is necessary for the parabola to exist.
karush
Gold Member
MHB
A stone is thrown vertically upward with a speed of 15.5 m/s
from the edge of a cliff 75.0 m high

ok, I was going to graph this with Desmos first
but don't know how to derive the equation for the parabola.

there are 11 questions following this...$$\tiny{Embry-Riddle \, Aeronautical \, University, \, HW3 \, PS103}$$

Last edited:
karush said:
A stone is thrown vertically upward with a speed of 15.5 m/s
from the edge of a cliff 75.0 m high

ok, I was going to graph this with Desmos first
but don't know how to derive the equation for the parabola.

there are 11 questions following this...

$f(t) = -\dfrac{1}{2}gt^{2} + v_{0}t + h_{0}$

$g$ is the Acceleration due to Gravity. Keep track of the sign and units.
$v_{0}$ is the Initial Velocity. Keep track of the sign and units.
$h_{0}$ is the Initial Height. Keep track of the sign and units.
$t$ is the time, in seconds.

P.S. Did I mention how important it is to keep track of the signs and units?

tkhunny said:
$f(t) = -\dfrac{1}{2}gt^{2} + v_{0}t + h_{0}$

$g$ is the Acceleration due to Gravity. Keep track of the sign and units.
$v_{0}$ is the Initial Velocity. Keep track of the sign and units.
$h_{0}$ is the Initial Height. Keep track of the sign and units.
$t$ is the time, in seconds.
\begin{align*} f(t)&=-\dfrac{1}{2}gt^{2}+v_{0}t+h_{0}\\ &=-0.5(9.8)t^2+15.5t+75 \end{align*}
ok I wasn't sure sure about $g$ but so far this
View attachment 7654
a. How much time will it take to reach the highest point?
\begin{align*} s(t)&=-\dfrac{1}{2}gt^{2}+v_{0}t+h_{0}\\ v(t)&=-9.8t+15.5 \\ v(1.58163)&=0 \end{align*}
well I got $\approx 1.582s$

#### Attachments

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Looks good.

It is important to remember that this model refers to the VERTICAL displacement ONLY. It is easy to mistake the pretty picture to mean something about horizontal displacement. We threw the object STRAIGHT UP and it falls STRAIGHT DOWN. Nothing horizontal in this model.

That is, the horizontal axis of your graph is time, not distance.

tkhunny said:
Looks good.

It is important to remember that this model refers to the VERTICAL displacement ONLY. It is easy to mistake the pretty picture to mean something about horizontal displacement. We threw the object STRAIGHT UP and it falls STRAIGHT DOWN. Nothing horizontal in this model.

however we are not graphing the actual path the object
but rhe speed of the objecr which is a parabola !

karush said:
however we are not graphing the actual path the object
but rhe speed of the objecr which is a parabola !

Speed? No. This is just vertical displacement - position.

You correctly observed where the velocity was zero (0) by finding the first derivative.

If you throw the stone in view of a fixed camera position, you would see no parabola - Just up and down.
If you repositioned the camera consistently and quickly in one direction, then you would see the parabola.

Vertical axis is vertical displacement (height).
Horizontal axis is time.

tkhunny said:
Vertical axis is vertical displacement (height).
Horizontal axis is time.

If it was not a parabola the t-axis
Would useless.

karush said:
If it was not a parabola the t-axis
Would useless.
I would tend to disagree and reverse your argument. Without the t-axis, it isn't a parabola. There isn't anything about throwing a rock straight up that says "parabola" UNLESS you watch the thing as it travels through time.

## 1. What is the initial velocity of the stone?

The initial velocity of the stone is 15.5 m/s.

## 2. What is the height of the cliff?

The height of the cliff is 75.0 m.

## 3. How long does it take for the stone to reach the top of the cliff?

The time it takes for the stone to reach the top of the cliff can be calculated using the formula t = v/g, where t is time, v is initial velocity, and g is the acceleration due to gravity (9.8 m/s^2). In this case, the time it takes for the stone to reach the top of the cliff is approximately 1.6 seconds.

## 4. What is the maximum height reached by the stone?

The maximum height reached by the stone can be calculated using the formula h = v^2/2g, where h is the maximum height, v is initial velocity, and g is the acceleration due to gravity. In this case, the maximum height reached by the stone is approximately 12.6 meters.

## 5. What is the total time the stone is in the air?

The total time the stone is in the air can be calculated by doubling the time it takes for the stone to reach the top of the cliff. In this case, the total time the stone is in the air is approximately 3.2 seconds.

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