A car is parked on a cliff overlooking the ocean on an incline that makes an ang

Question

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 16.0° 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 2.59 m/s2 for a distance of 65.0 m to the edge of the cliff, which is 40.0 m above the ocean.

(a) Find the car's position relative to the base of the cliff when the car lands in the ocean.
_________m

(b) Find the length of time the car is in the air.
_________s

Explanation / Answer

To find the velocity as it reaches the end of the cliff, first we need the time it spends rolling down the slope. We can solve the position equation for time to do this: x(t) = x0 + v0*t + (1/2)*a*t^2 65 = 0 + 0 + (1/2)*2.59*t^2 t = 7.085 seconds meaning v = t*a = 7.085*2.59 = 18.35 m/s at 16 degrees below horizontal if we call the edge of the cliff the origin, and say the ocean is in the positive x direction.. thus: Vx = 18.35*cos(-16) = 17.639 Vy = 18.35*sin(-16) = -5.058 Projectile motion in the Y, for a drop of 40m, to solve for t when it hits the ocean (also equal to and answer for part b): y(t) = y0 + Vy - (1/2)*g*t^2 -40 = 0 - 5.058*t - (1/2)*9.81*t^2 t = 2.386 seconds (answer to part b) x(t) = Vx*t = 17.639*2.386 = 42.087 m from the base of the cliff (answer to part a)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.