Problem in equations of kinematics in two dimensions

[Batlin]

Problem in equations of kinematics in two dimensions..please help

Hi, I'm a first year physics student and am studying equations of kinematics in two dimensions and projectile motion. I was given the following question as homework to prep for my first exam this Wednesday. I'm pretty solid on the subject but for some reason I am stuck on this problem:

Homework Statement

"A car drives straight off the edge of a cliff that is 54m high. The point of impact is 130m from the base of the cliff. How fast was the car traveling when it went off the cliff?"

Homework Equations

Vx = Vox + AxT
X = 1/2(Vox + Vx)T
X = VoxT + 1/2AxT^2
Vx^2 = Vox^2 + 2AxX

Vy = Voy + AyT
Y = 1/2(V0y + Vy)T
Y = VoyT + 1/2AyT^2
Vy^2 = Voy^2 + 2AyY

The Attempt at a Solution

Based on the information given, I know my Delta X and Delta Y. However, no matter which equation I use to sovle for Vox (the speed of the car when it left the cliff) I always have 2 unknown values in my equation. If I knew Vx or T I could solve this. However, I know that it is solvable, my professor posted T and VoX. I am just totally stuck on where to take the first step in the problem.

Any help is greatly appreciated.
Bat.

 

[LowlyPion]

Hi, I'm a first year physics student and am studying equations of kinematics in two dimensions and projectile motion. I was given the following question as homework to prep for my first exam this Wednesday. I'm pretty solid on the subject but for some reason I am stuck on this problem:

Homework Statement

"A car drives straight off the edge of a cliff that is 54m high. The point of impact is 130m from the base of the cliff. How fast was the car traveling when it went off the cliff?"

Homework Equations

Vx = Vox + AxT
X = 1/2(Vox + Vx)T
X = VoxT + 1/2AxT^2
Vx^2 = Vox^2 + 2AxX

Vy = Voy + AyT
Y = 1/2(V0y + Vy)T
Y = VoyT + 1/2AyT^2
Vy^2 = Voy^2 + 2AyY

The Attempt at a Solution

Based on the information given, I know my Delta X and Delta Y. However, no matter which equation I use to sovle for Vox (the speed of the car when it left the cliff) I always have 2 unknown values in my equation. If I knew Vx or T I could solve this. However, I know that it is solvable, my professor posted T and VoX. I am just totally stuck on where to take the first step in the problem.

Any help is greatly appreciated.
Bat.

Welcome to PF.

You know the height, so you can calculate the time to impact.

That times the horizontal velocity is known to be 130 m.

 

[Batlin]

Hrmm.. see this is where I'm confused. I only have a distance. I don't know my initial or ending velocity or the time it took to travel the 130m?

 

[Batlin]

Welcome to PF.

You know the height, so you can calculate the time to impact.

That times the horizontal velocity is known to be 130 m.

I did try this route once:

130m = 54mT, T = 2.41

However, the professor gave the answer to the problem and has T = 3.32 sec

I cannot figure out how she got this?

 

[LowlyPion]

Hrmm.. see this is where I'm confused. I only have a distance. I don't know my initial or ending velocity or the time it took to travel the 130m?

But you have more information that is not stated.

You have gravity. It's easy to forget because it's always there.

How long does it take the car to drop vertically in the gravity field off the cliff?

That would be given by H = 54 m = 1/2 (9.8 m/s²) (t²)

You can find these equations in your textbook or from this link:
https://www.physicsforums.com/showpost.php?p=905663&postcount=2

 

[LowlyPion]

I did try this route once:

130m = 54mT, T = 2.41

However, the professor gave the answer to the problem and has T = 3.32 sec

I cannot figure out how she got this?

This equation just can't be right.

You have distance = distance * time.

The ratio of the distances in this case is simply not the time.

 

[Batlin]

But you have more information that is not stated.

You have gravity. It's easy to forget because it's always there.

How long does it take the car to drop vertically in the gravity field off the cliff?

That would be given by H = 54 m = 1/2 (9.8 m/s²0 (t²)

I see now!

I get 3.3197 sec for T.

Damn how did I miss this... what's screwing me up is that I kept thinking I needed a Vo or V to get my Time.

 

[LowlyPion]

I see now!

I get 3.3197 sec for T.

Damn how did I miss this... what's screwing me up is that I kept thinking I needed a Vo or V to get my Time.

Here is a link with common kinematic formulas.
https://www.physicsforums.com/showpost.php?p=905663&postcount=2

You look good to go now.

Good Luck.