Question

The end of an action movie car chase involves the villain driving off a vertical cliff and falling into the ocean below You have been asked to work out the details of this stunt. The diff is 100m above the water, and the water begins 30m away from the base of the cliff. The car is supposed to land at least 90m Into the water. The edge of the cliff has a slight upward slope, 15 degrees above horizontal. Draw a motion diagram and a force diagram for the car while it is airborne Estimate how fast the car should be going when it leaves the end of the cliff. Do you expect this to be an underestimate or an overestimate? Explain your reasoning a. b.

Answer 1

Answer:

The initial velocity should be greater than 23.9 m/s. This velocity is underestimated because it does not take into account the resistance of the air.

Explanation:

Hi there!

The final position vector of the car (denoted as "r" in the figure) is

r = ( 120, -100) m

(placing the origin of the frame of reference at the launching point).

The position vector at a time "t" is calculated by the following equation:

r = (x0 + v0 · t · cos θ, y0 + v0 · t · sin θ + 1/2 · g · t²)

Where:

x0 = initial vertical position.

v0 = initial velocity.

t = time.

θ = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity.

At final time:

120 m = x0 + v0 · t · cos θ

-100 m = y0 + v0 · t · sin θ + 1/2 · g · t²

Since the origin is placed at the launching point, x0 and y0 = 0, then:

120 m = v0 · t · cos θ

-100 m = v0 · t · sin θ + 1/2 · g · t²  

We have a system with two equations and two unknowns, so, we can solve it.

Solving the first equation for "v0":

120 m = v0 · t · cos θ

v0 = 120 m / (cos θ · t)

Replacing v0 in the second equation:

-100 m = v0 · t · sin θ + 1/2 · g · t²  

-100 m = (120 m/cos 15° · t) · t · sin 15° - 1/2 · (9.8 m/s²) · t²  

-100 m = 120 m · tan 15° - 4.9 m/s² · t²

(-100 m - 120 m · tan 15°) / (-4.9 m/s²) = t²

t = 5.2 s

Now we can calculate the initial velocity:

v0 = 120 m / cos 15° · 5.2 s

v0 = 23.9 m/s

The initial velocity should be greater than 23.9 m/s

This velocity is underestimated because it does not take into account the resistance of the air.