Question

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 25.0

Answer 1

B) 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*3.97*t^2
and
v=3.97*t
solve first for the t, which is the rolling time

40=.5*3.97*t^2
t=4.489 seconds
so v=17.8213 m/s
the velocity has direction, which is 25 degrees below the horizontal.

Once airborne, the car has the vertical equation of motion for position of
y(t)=35-17.8213*sin(25)*t-.5*9.81*t^2
when y(t)=0, splashdown

4.905*t^2 + 7.5316*t - 35
when t=2.013 seconds, the car hits the water. That is the time in the air

A) The horizontal equations of motion are
x(t)=17.8213*cos(25)*t
to find the distance from the base of the cliff
x(2.27)=17.8213*cos(25)*2.013

32.51313 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= -3.544 seconds