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Cliff Projectile Motion

by jsr219
Tags: cliff, motion, projectile
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jsr219
#1
Sep17-11, 07:50 PM
P: 8
Your friends car is parked on a cliff overlooking the ocean on an incline that makes an angle of 18.9 (degrees) below the horizontal. The brakes fail, and the car rolls from rest down the incline for a distance of 26.7 m to the edge of the cliff, which is 82.4 m above the ocean, and, unfortunately, continues over the edge and lands in the ocean.

Find the cars position relative to the base of the cliff when the car lands in the ocean.




I'm not quite sure how to attempt this problem
- I think you bust the problem up into 2 parts
- I know you have to find the velocity of the car right before it falls off the cliff. Then from there treat it as a projectile question.

- I also know that I will have to break the problem down into its x and y components


If you could help walk me through this problem it would be much appreciated
- I deff don't want anyone to give a final answers
- just help walk me through the problem step by step so that I can understand whats going on

I just have no clue how to get started or the equations to use
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Liquidxlax
#2
Sep17-11, 08:17 PM
P: 321
You want to start by calculating the force normal and the force parallel to the incline, so you can get the accelerations. You will also want to calculate the change in x and change in y on the incline. So say the car starts at the origin (0,0) and at the end of the cliff is (x, y). You know the hypotenus is 26.7m and it is a 18.9 degree slope

It is important to write down all the values you are given and the equations which include some of your variables

x = 26.7cos(18.9)

y = 26.7sin(18.9)

ay = gcos(theta)

ax = gsin(theta)

d = vi(t) + .5at^2

at = v

You should know your kinematic equations

and solve the equations for velocity on the ramp in the x and y directions

When the car leaves the cliff there is no more acceleration in the x direction so the x velocity becomes constant, but in the y direction gravity is acting upon the car. So there is a higher constant acceleration in the y direction which is 9.81m/s^2

You know the height of the cliff the initial y velocity after it leaves the cliff and the acceleration. So you must find the time it takes for the car to reach the ocean. You can do this by using the equation above and the quadratic equation with respect to t. You should be able to see which time it will be.

Then use the x velocity at the end of the cliff and the time you've found for the car to reach the ocean from the end of the cliff and multiply them together to find the distance the car traveled from the cliff.

hope this helps


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