(a) The car landed 29.9 m from the base of the cliff. Position relative to the base of the cliff: r = (29.9 m, 0)
(b) The car was in the air for 1.76 s
Explanation:
Please see the attached figure for a graphical description of the problem.
In this problem, we have two motions. The rolling motion, which is an accelerated motion in a straight line, and the falling motion, that is a semi-parabolic motion.
These are the equations we are going to use:
1) The position and velocity of the car while rolling down:
x =x0 + v0 * t + 1/2 a * t²
v = v0 + a * t
where:
x = position of the car at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
2) The vectors of the position and velocity of the car while falling:
r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)
v = (v0x, v0y + g * t)
Where:
r = position vector
x0 = initial horizontal position
v0x = initial horizontal velocity
t = time
y0 = initial vertical position
v0y = initial vertical velocity
g = acceleration due to gravity
v= velocity vector
a) The position of the car relative to the base of the cliff when the car lands is the x-component of the vector "r final" (see figure).
r final x = x0 + v0x * t (see the equation of the x-component of "r" above)
To find this distance, we need to know the initial horizontal velocity and the time at which the car lands.
Seeing the figure, notice that the the vectors v0x, v0y and v0 form a right triangle. Then, using trigonometry:
cos 18.0º = adjacent / hypotenuse = v0x / v0
sin 18.0º = opposite / hypotenuse = v0y / v0
But now, we still do not have v0.
The initial velocity of the falling car is the velocity of the rolling car when it reaches the edge of the cliff. This velocity can be calculated once we have the time at which the car reaches the edge. From the equation for position:
x =x0 + v0 * t + 1/2 a * t²
Since the origin of the reference system is located at the point at which the car starts rolling, x0 =0. The car starts rolling from rest, then, v0 = 0.
Therefore:
x = 1/2 a * t²
55.0 m = 1/2 * 2.90 m/s² * t²
t² = 2 * 55.0 m / 2.90 m/s²
t = 6.16 s
The velocity at the edge of the cliff will be:
v = a * t (since v0 = 0)
v = 2.90 m/s² * 6.16 s = 17.9 m/s
Now, we can calculate v0x and v0y:
cos 18.0º = v0x/v0
v0* cos 18.0º = v0x
v0x = 17.9 m/s * cos 18º = 17.0 m/s
sin 18.0º = v0y / v0
v0y = v0 * sin 18.0º = 17.9 m/s * sin 18.0º = 5.53 m/s (notice that this is the magnitude of v0y. Since its direction is downward, its sign must be negative).
Now, using the equation for the y-component of the vector "r final", we can calculate how much time the car was falling:
y = y0 + v0y * t + 1/2 * g * t²
Let´s place now the center of our reference system at the point at which the car starts falling. Then y0 = 0. For the vector "r final" (position of the car at final time) the y-component is -25.0 m (see figure). Then:
-25.0 m = v0y * t + 1/2 * g * t²
-25.0 m = -5.53 m/s * t + 1/2 * (-9.8 m/s²) * t²
-25.0 m = -5.53 m/s * t - 4.9 m/s² * t²
0= 25.0 m - 5.53 m/s * t - 4.9 m/s² * t²
Solving the quadratic equation:
t = 1.76 s and t = -2.89 (this value is discarded)
Now, having the time at which the car lands and the initial horizontal velocity, we can calculate at which distance from the base of the cliff the car landed:
The x-component of r final is
x = x0 + v0x * t (since the edge of the cliff is the origin of our reference system x0 = 0).
x = v0x * t = 17.0 m/s * 1.76 s = 29.9 m
The car landed 29.9 m from the base of the cliff. Relative to the cliff, the vector position will be r = (29.9 m, 0). Relative to the edge, the position vector will be r = (29.9 m, -25.0 m)
b) This time was already calculated using the y-component of the vector r final. We know that at the final time the y-component of that vector is -25.0 m. Using that and v0y, we could obtain a quadratic equation. The solution of the equation was the total time the car was in the air: 1.76 s ( see above).
