Ask your homework questions to teachers and professors, meet other students, and be entered to win $600 or an Xbox Series X 🎉Join our Discord!



Carnegie Mellon University



Problem 15 Medium Difficulty

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of $24.0^{\circ}$ below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 4.00 $\mathrm{m} / \mathrm{s}^{2}$ for a distance of 50.0 $\mathrm{m}$ to the edge of the cliff, which is 30.0 $\mathrm{m}$ above the ocean. Find (a) the car's position relative to the base of the cliff when the car lands in the ocean and (b) the length of time the car is in the air.


(a) $d = 32.52 \mathrm { m }$
(b) $t = 1.78 \mathrm { s }$


You must be signed in to discuss.

Video Transcript

so the speed of the car when it reaches the edge of the cliff, this will be the final, equaling the square root of the initial squared, which we know to be zero. Because it's starting from rest plus two times the acceleration times, the horizontal displacement and so this would be equal in the square root of two times 4.0 meters per second squared multiplied by 50 meters, 50.0 meters. Extend the square root, and we find that this is equaling 20.0 meters per second. Now we can, uh, take the car to be a projectile once it fault wants it once it drives off the edge of the cliff than the only force that the car is under is, of course, the force of gravity. Which means that the car is in free fall. And so we can now then say that the vertical velocity of the car as it reaches the water we will say that the final here, we'll save the Y final. This would be equaling the negative square root will say negative because we know the car is traveling downwards. Um ah, negative one, essentially times the square root of velocity. Why initial squared, plus two times the acceleration of the Y direction times delta y. We know that we can then say that the Y final is equally negative and then this would be the square root of negative 20 0.0 meters per second. This will be time sign of 24 degrees and this will be squared plus two times negative 9.80 meters per second squared multiplied by negative 30.0 meters. So I would be the vertical displacement. And we find that then v y final is equaling negative 25 0.6 meters per second and so we can find the time of flight for part a rather for part B. And you can say that then the time of flight would be equaling. Be why final minus and B. Why initial which we know this. Rather, this would be divided by the acceleration lower direction. This is equaling negative 25.6 meters per second minus and then this would be negative 20 meters per second, multiplied by sine of 24 degrees, and this would be all divided by negative 9.80 meters per second squared and the time is equaling 1.78 seconds. So this would be our answer for part B and then for part a. The horizontal displacement of the car delta acts would be equaling the X final multiplied by t and so this would be equaling 20.0 meters per second. We don't have to account for the acceleration term because there isn't any acceleration in the ex direction. So that would be 20.0 meters per second. Most applied by co sign of 24 degrees. And this would be multiplied by again 1.78 seconds. And we find the our horizontal displacement is equaling 32.5 meters. This would be our final answer. That is the end of the solution. Thank you for watching.

Carnegie Mellon University
Top Physics 101 Mechanics Educators
Elyse G.

Cornell University

Christina K.

Rutgers, The State University of New Jersey

Liev B.

Numerade Educator

Aspen F.

University of Sheffield