Answer:
a) the car lands in the ocean 85.1 meters away from the base of the cliff.
b) the length of time the car is in the air is 2.50 seconds.
Step-by-step explanation:
Let's start with part (a) of the problem.
First, we need to find the car's velocity at the bottom of the incline using the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity (which is 0 m/s), a is the acceleration (which is 3.67 m/s^2), and s is the distance traveled down the incline (which is 50.0 m).
Plugging in these values, we get:
v^2 = 0^2 + 2(3.67 m/s^2)(50.0 m)
v^2 = 367 m^2/s^2
v = 19.1 m/s (rounded to one decimal place)
Next, we can use the vertical motion equations to find the time it takes for the car to fall from the cliff to the ocean. We'll use the equation:
h = ut + (1/2)at^2
where h is the height of the cliff (35.0 m), u is the initial vertical velocity (which is 0 m/s), a is the acceleration due to gravity (-9.81 m/s^2), and we're solving for t.
Plugging in these values, we get:
35.0 m = 0 m/s * t + (1/2)(-9.81 m/s^2)t^2
19.9 = t^2
t = 4.46 s (rounded to two decimal places)
Therefore, the car is in the air for 4.46 seconds.
Finally, to find the car's position relative to the base of the cliff when it lands in the ocean, we can use the horizontal motion equation:
s = ut + (1/2)at^2
where s is the horizontal distance the car travels (which is what we're solving for), u is the horizontal velocity (which is the same as the velocity at the bottom of the incline, 19.1 m/s), a is the horizontal acceleration (which is 0 m/s^2), and t is the time the car is in the air (which is 4.46 s).
Plugging in these values, we get:
s = 19.1 m/s * 4.46 s + (1/2)(0 m/s^2)(4.46 s)^2
s = 85.1 m (rounded to one decimal place)
Therefore, the car lands in the ocean 85.1 meters away from the base of the cliff.