# Kinematics Equations and a Cliff

• MarchON
In summary, the conversation discusses a problem where a car drives off a 100m cliff and needs to land 90m into the water that starts 30m away from the cliff. The goal is to find the speed of the car in order to land at the desired point in the water, assuming negligible air resistance. The conversation covers the use of kinematics and the need to consider the horizontal and vertical components of the car's motion. It also discusses the incorrect assumption of air resistance being negligible and its impact on the final answer.
MarchON

## Homework Statement

A car drives off a cliff that is 100m high. It has to land in water and the water starts 30m away from the cliff. Its goal is to land 90m into the water. How fast must the car be going to land at that point in the water. Air resistance is negligible.

v=0
v0=?
a= -9.91m/s2
Δy=100m
Δx=120m

## Homework Equations

My plan was to use kinematics and determine the time it takes for the car to fall using Δy= vt - 1/2(a)t2, then put that into Δx= 1/2(v+v0)t. Isn't it true that whether you throw something horizontally or just drop it, it takes the same time to hit the ground? Why does that not apply here regarding the horizontal distance and velocity?

## The Attempt at a Solution

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My answer is wrong. I get 53 m/s (this seems like a lot anyway).

MarchON said:
My plan was to use kinematics and determine the time it takes for the car to fall using Δy= vt - 1/2(a)t2, then put that into Δx= 1/2(v+v0)t.
Think over the last part of your strategy, keeping in mind a condition from the initial problem statement.
MarchON said:
Air resistance is negligible.

The strategy is certainly correct. I get a much smaller answer, though.
I find your horizontal motion distance formula confusing with the two velocities and division by 2.
The horizontal velocity is constant; why not just call it v? Try running it through that way!

MarchON said:

## Homework Statement

A car drives off a cliff that is 100m high. It has to land in water and the water starts 30m away from the cliff. Its goal is to land 90m into the water. How fast must the car be going to land at that point in the water. Air resistance is negligible.

v=0
v0=?
a= -9.91m/s2
Δy=100m
Δx=120m

a=-9.81 m/s2
That means you took upward positive. Will the car move upward when falling from the cliff? What is the sign of Δy?

MarchON said:

## Homework Equations

My plan was to use kinematics and determine the time it takes for the car to fall using Δy= vt - 1/2(a)t2, then put that into Δx= 1/2(v+v0)t. Isn't it true that whether you throw something horizontally or just drop it, it takes the same time to hit the ground? Why does that not apply here regarding the horizontal distance and velocity?

Δy= vt - 1/2(a)t2 is true if v means the vertical component of the initial velocity. But the car drives on a horizontal cliff. What is the vertical component of velocity at the instant when it leaves the cliff??
Δx= 1/2(v+v0)t is valid for uniform accelerating motion in the x direction, (horizontally). Is there any horizontal force applied on the car, causing acceleration in the horizontal direction, and changing the horizontal component of velocity?And you can not use the same notation v for the initial vertical velocity and the final horizontal velocity components.

It is true that whether you throw something horizontally or just drop it, it takes the same time to hit the ground.

First of all whether you throw something horizontally or dropping it does take the same time. However this does not mean it goes the same horizontal distance, the x-component of the velocity determines that. Also there is no velocity in the y-direction as the car is moving horizontally. What you are going to have to do is find the time it takes the car to hit the ground using
y = y0-1/2*g*t^2, Where g = 9.8 m/s^2 and y0 is the initial height.
Then use ∆x = v0t. Use the time (t) you solved for in the y equation and plug it into this x equation to solve for v0.

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Bystander said:
Think over the last part of your strategy, keeping in mind a condition from the initial problem statement.

Air resistance being negligible is part of the problem and becomes in incorrect assumption for part C. It's there to show us that the final answer, whatever it is, will be inaccurate... I am having trouble getting to that final answer.

I don't see any parts in the question posted. What is part C?

@velo city: The policy of the forum does not allow to solve the problem for the OP. And it is even worse to show wrong solution, as yours is for the time for the fall. Also your formula for the ∆x is misleading as the OP used the notation a for -9.81.

ehild said:
Δy= vt - 1/2(a)t2 is true if v means the vertical component of the initial velocity. But the car drives on a horizontal cliff. What is the vertical component of velocity at the instant when it leaves the cliff??
Δx= 1/2(v+v0)t is valid for uniform accelerating motion in the x direction, (horizontally). Is there any horizontal force applied on the car, causing acceleration in the horizontal direction, and changing the horizontal component of velocity?And you can not use the same notation v for the initial vertical velocity and the final horizontal velocity components.

I thought that v was v final which is 0 with regards to the vertical portion of its fall? I thought v0 was the initial velocity which is what I am trying to find.

Delphi51 said:
I don't see any parts in the question posted. What is part C?

I did not post that part because I understood it.
I was just letting you know, so people don't try to factor in air resistance mathematically.

ehild said:
@velo city: The policy of the forum does not allow to solve the problem for the OP. And it is even worse to show wrong solution, as yours is for the time for the fall. Also your formula for the ∆x is misleading as the OP used the notation a for -9.81.
sorry i will change it.

The v and vo are in the horizontal motion formula with no acceleration. We used to write d = vt in high school. This v is the speed of the car you are trying to find.

MarchON said:
I thought that v was v final which is 0 with regards to the vertical portion of its fall? I thought v0 was the initial velocity which is what I am trying to find.
The vertical and horizontal motions of the car are independent if air resistance is ignored. The vertical motion is free fall. The displacement is Δy= vy0t - 1/2(g)t2
where vy0 is the initial vertical velocity. What is its value?
The velocity of the horizontal motion does not change during the fall. It is the unknown v.

ehild said:
@velo city: The policy of the forum does not allow to solve the problem for the OP. And it is even worse to show wrong solution, as yours is for the time for the fall. Also your formula for the ∆x is misleading as the OP used the notation a for -9.81.
By the way I was using the time for the fall to find the x displacement. So my solution was not wrong.

ehild said:
The vertical and horizontal motions of the car are independent if air resistance is ignored. The vertical motion is free fall. The displacement is Δy= vy0t - 1/2(g)t2
where vy0 is the initial vertical velocity. What is its value?
The velocity of the horizontal motion does not change during the fall. It is the unknown v.
I don't think the car has an initial vertical velocity right?

ehild said:
where vy0 is the initial vertical velocity. What is its value?
0?

Anyway, I have the solution in front of me. It has a bunch of trig which I do not understand. The solution used the "standard coordinate system with the origin at the initial position of the car."

x = x0 + (v0,x)t+(1/2)(ax)t2 = 0 + v0(cosθ)t+(1/2)(0)t2=v0(cosθ)t

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MarchON said:
0?

Anyway, I have the solution in front of me. It has a bunch of trig which I do not understand. The solution used the "standard coordinate system with the origin at the initial position of the car."

x = x0 + (v0,x)t+(1/2)(ax)t2 = 0 + v0(cosθ)t+(1/2)(0)t2=v0(cosθ)t
the trig comes only from projectiles launched at an angle. In this case that angle (ø) is 0 because the car is only moving horizontally.

velo city said:
the trig comes only from projectiles launched at an angle. In this case that angle (ø) is 0 because the car is only moving horizontally.
You should probably watch this video: . It will help you understand if you watch the whole thing. I don't want to give you a solution cause that is against the rules. So that video should help.

Also watch this one:

velo city said:
By the way I was using the time for the fall to find the x displacement. So my solution was not wrong.
The time (more thn 20 s) was wrong. (You removed it already ). Also, the notation was misleading in the formula of x displacement. What value for a did you use? As the OP set a=-9.81 m/s2.

I used a =0 as there is no acceleration in the x direction

I also forgot to take the square root and divide when I solved for the time. Stupid me.

velo city said:
I used a =0 as there is no acceleration in the x direction
That is correct.
One has to use different notations for different things, and also take attention of the notations used by the OP in the thread. ax instead of a here.

velo city said:
I also forgot to take the square root and divide when I solved for the time. Stupid me.
Yes :) You see, it is much safer not to give out numerical solutions.

Ok, well my number has gotten smaller, but it is still not the 23.9 m/s that is apparently the answer.

I solved for time and got 4.5 seconds, and then using Δx=v0t+(1/2)(a)t2

I got 26.7 m/s

What value did you use for a?

0 because it's the x direction.

Right?

ehild said:
What value did you use for a?
Right. Also the result v=26.7 m/s for the horizontal velocity was right.

The answer my professor has given is 23.9 m/s.

MarchON said:
The answer my professor has given is 23.9 m/s.
show him that he was wrong :)

ehild said:
show him that he was wrong :)

Haha, not worth it. And I have no idea why he made the solution so overly confusing.

MarchON said:
Haha, not worth it. And I have no idea why he made the solution so overly confusing.
Coward! A brave man should fight for the truth!
You can ask him to explain the "correct" solution to you as you got different result... And then he either founds his own error or we find his error or our error, if any.
If he made the solution confusing, he could have confused himself.

I really like the two headings for horizontal and vertical parts of the motion in that example. That is a great first step in any two dimensional motion problem. The second step is to identify which part is accelerated motion and write the appropriate formulas.

## 1. What are the kinematics equations used to describe motion in a cliff?

The kinematics equations used to describe motion in a cliff are the equations of motion, also known as the SUVAT equations. These equations are:

1. v = u + at (equation for final velocity)

2. s = ut + 1/2at^2 (equation for displacement)

3. v^2 = u^2 + 2as (equation for final velocity squared)

4. s = (u + v)/2 * t (equation for average velocity)

5. v = u + 2a(s - u)/t (equation for final velocity with displacement and time)

## 2. How do these equations apply to motion in a cliff?

These equations can be used to describe the motion of an object falling off a cliff. The initial velocity (u) would be zero, as the object starts from rest at the top of the cliff. The acceleration (a) would be equal to the acceleration due to gravity (9.8 m/s^2). The displacement (s) would be the height of the cliff, and the time (t) would be the time it takes for the object to reach the ground.

## 3. Can these equations be used for any type of motion in a cliff?

Yes, these equations can be used for any type of motion in a cliff, as long as the acceleration remains constant. This means that the object is moving in a straight line with a constant acceleration, such as in free fall.

## 4. What is the significance of using these equations in studying motion in a cliff?

These equations allow us to accurately predict the motion of an object falling off a cliff. By plugging in known values, we can calculate the final velocity, displacement, and time taken for the object to reach the ground. This can be useful in understanding the physics of falling objects and in predicting the outcome of real-life scenarios, such as a skydiver jumping off a cliff.

## 5. Are there any limitations to using these equations for motion in a cliff?

These equations assume that the acceleration remains constant, which may not always be the case in real-life scenarios. For example, air resistance can affect the acceleration of an object falling off a cliff. Additionally, these equations do not take into account other factors such as the shape or mass of the object, which can also affect its motion. Therefore, while these equations can provide a good estimate, they may not always accurately describe the motion in a cliff in real-life situations.

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