PHYSICS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?

: PHYSICS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?
A car is parked on a cliff overlooking the ocean
on an incline that makes an angle of 16.6◦
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 6.24 m/s^2 and travels 31.9 m to the edge of
the cliff. The cliff is 15.9 m above the ocean.
The acceleration of gravity is 9.8 m/s^2

A) Find the car’s position relative to the base
of the cliff when the car lands in the ocean.
Answer in units of m

B)Find the length of time the car is in the air.
Answer in units of s

Answers and Views:

Answer by civil_av8r
Find the speed of the car before it leaves the cliff

vf^2 = vi^2 + 2*a*d

vf = sqrt(vi^2 + 2*a*d)
and since the car starts from rest, vi=0, vf = sqrt(2*a*d)

Find the initial speed in the vertical direction, vi
vi = vf*sin(-16.6) = sqrt(2*a*d)*sin(-16.6)

Equation of motion in the vertical direction
hf = hi + vi*t – 1/2*g*t^2

hf = 0
hi = 15.9 m
vi = sqrt(2 * 6.24 m/s^2 * 31.9 m)*sin(-16.6)

0 = hi + vi*t – 1/2*g*t^2

Use the quadratic equation to find the values of t. More than likely one number will be negative so you can immediately throw that one out and use the other one. This is your answer for part (B).

After you find ‘t’ simply use the equation of motion in the horizontal direction to find the distance.

s = s0 + vi*t + 1/2*a*t^2

In the horizontal direction:
s0 = 0
vi = v*cos(-16.6) = sqrt(2*a*d)*cos(-16.6)
a = 0

s = 0 + sqrt(2*a*d)*cos(-16.6)*t + 0 <---- (A) Plug and chug.