# Mechanics problem confusing!

1. Feb 26, 2010

### luysion

1. The problem statement, all variables and given/known data
"A cat leaps horizontally from a flat roof 7.5 m above the ground and lands on another flat roof 5 m above the ground and 3.5 m away horizontally.

How long, in seconds, does the leap take?"

2. Relevant equations

3. The attempt at a solution
Ok, i imagined the situation as half a projectile. with the cat travelling the hypotenuse of the triangle that can be formed (2.5 and 3.5 base). I know vi = 0 at highest point of a projectile and i know that the distance is 4.3 (pythagoras) i used kinematics ( d = vit + 1/2at^2) and get an answer of 0.81..

CHeers

2. Feb 26, 2010

### aftershock

Remember that the horizontal and vertical components are completely seperate. If you focus on just vertical components, you should see that you have the values for acceleration, original velocity, and displacement.

Use kinematics formulas to solve for time.

3. Feb 26, 2010

### luysion

I did that, and I didnt get 0.7 as my answer. I calculated the vertical component as 4.3 so thats my d(vertical) and i know vi(vertical component) is zero at the highest point of a projectile and acceleration is 9.8..

what am I doing wrong???

4. Feb 26, 2010

### songoku

Hi luysion

This is your mistake. The trajectory of the cat is not straight line but parabolic so it's nothing to do with triangle. Just use simple kinematics as aftershock suggested. Consider the vertical components to find the time needed

5. Feb 26, 2010

### aftershock

There's an acceleration so you need an initial and final velocity. The problem said the cat leaps horizontally... so there would be no vertical velocity at all to begin with. Displacement would be the difference between the two roofs.

Post what number's you're using for original velocity and displacement... and then what equation you're using if you still can't get .7 :)

6. Feb 26, 2010

### Kommandant

This is how you solve it.

There are two building, the height difference between both of them is 2.5 meters.

Y (displacement in the vertical axis) = 2.5

We are in Earth, so the acceleration due to gravity is 9.81 m/s^2

a (acceleration) = 9.81

The Cat is resting and then begins descending; therefore, his initial speed is 0 m/s

Vi = 0 m/s

Solve

Y=ViT+(1/2)at^2

2.5=(1/2)(9.81)t^2

(5/9.81)=T^2

Square Root of (5/9.81) = .7139

7. Feb 27, 2010

### luysion

Ah Cheers guys! thanks for all the replies. I was using the wrong component! i forgot in projectile we use the vertical component in kinematic equations. I kept using the hypotenuse which made no sense -_- as a projectile is a parabolic path

thanks so much to everyone!!1