Calculating Velocity of a Car Rolling Off a Cliff

[loganblacke]

Homework Statement

A car is parked on an incline of 15.6 degrees below the horizontal, the brakes fail and it rolls down the incline 27.5m and off the edge of a cliff that is 57m above the ground. Find the car's position when it its the ground relative to the base of the cliff.

Homework Equations

rather than type them all out, I'll just say kinematics.

The Attempt at a Solution

I know you have to break the gravity vector up into components so I took -9.8 sin(15.6) = -2.63 m/s^2. I keep calculating the average speed of the car, not the speed at the point it goes off the cliff. Please help, thanks!

 

[cosmo123]

v^2=u^2+2as, so the car is traveling at (2*9.8sin(15.6)*27.5)^0.5 when it goes off the cliff.

 

[verty]

You have not said enough. You keep calculating the average speed. Do you want to calculate the average speed? Could you use the average speed? What is your plan for this question?

 

[loganblacke]

v^2=u^2+2as, so the car is traveling at (2*9.8sin(15.6)*27.5)^0.5 when it goes off the cliff.

Thanks, but I'm still getting the wrong answer. I end up with the velocity of the car being 12.04 m/s when it leaves the cliff. So initial velocity in the y direction when the car leaves the cliff should be 12.04*sin(15.6) = 3.24. Plug that into y=Vi(t)-4.92(t)^2 to solve for t. End up with 57 = 3.24(t) - 4.92(t)^2.. rewritten in quadratic form.. 4.92t^2-3.24t-57=0. then t = 3.748 seconds which is not correct.

 

[loganblacke]

You have not said enough. You keep calculating the average speed. Do you want to calculate the average speed? Could you use the average speed? What is your plan for this question?

How is that not enough information? You have a car rolling down a slope then off a cliff. You obviously need to know the velocity of the car as it rolls of the cliff to answer the question. I calculated the acceleration but couldn't remember how to get Vfinal.