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 20 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 2.4 m/s2 and travels 53 m to the edge of the cliff. The cliff is 35 m above the ocean. How long is the car in the air? The acceler-
ation of gravity is 9.81 m/s2 . Answer in units of s
What is the car’s position relative to the base of the cliff when the car lands in the ocean? Answer in units of m

Explanation / Answer

First thing we need to do is solve for the car's velocity the instant that it leaves the cliff(assuming it doesn't scrape or rock on the edge for a little while)
v2=v02+2ad

v=(0+2(2.4m/s)(53m))

v=11.28 m/s @ -200

To solve for flight time we only use the verticle component of speed

vy=v(sin)

=11.28(sin20)

vy=3.857 m/s

Plug this speed and the height into the equation where a is g=-9.81m/s2

h=h0+v0t+(1/2)at2

0=(35m) + (3.857m/s)t + (1/2)(-9.81m/s2)t2

Solve for t (you'll have to brush up on your quadratic formula)

In order to solve for distance traveled remember that once the car is in freefall the horizontal speed is constant so you can use the simple equation

d=vt

where v is vx=v(cos20)

and t is the time you solve for above

Hope this helps. Message me if you're confused about anything up here

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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