Physics problem- Car on cliff- please help!!!!!?

Question

A.) 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 4.10 m/s^2 and travels 43 m to the edge of the cliff. The cliff is 37m above the ocean. The acceleration of gravity is 9.81 m/s^2.
How long is the car in the air? Answer in units of s.

B.) 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.

Answer 1

First step is to determine the speed of the car at the moment it leaves the cliff.

The equations of motion while the car is rolling are:

d=.5*4.10*t^2
and
v=4.10*t
solve first for the t, which is the rolling time

43=.5*4.10*t^2
t=4.58 seconds
so v=18.8 m/s
the velocity has direction, which is 16 degrees below the horizontal.

Once airborne, the car has the vertical equation of motion for position of
y(t)=37-18.8*sin(16)*t-.5*9.81*t^2
when y(t)=0, splashdown
when t=2.27 seconds, the car hits the water. That is the time in the air

B) The horizontal equations of motion are
x(t)=18.8*cos(16)*t
to find the distance from the base of the cliff
x(2.27)=18.8*cos(16)*2.27
41 m from the base

Note: In part A there was also a root for t that was negative. This is because if the car had been traveling in the parabolic path throughout the airborne motion, it would have been at y=0 when t= -3.32 seconds

j