Your friend's car is parked on a cliff overlooking the ocean on an incline that makes an angle of \(17.0^{\circ}\) below the horizontal. The brakes fail, and the car rolls from rest down the incline for a distance of \(29.0 \mathrm{~m}\) to the edge of the cliff, which is \(55.0 \mathrm{~m}\) above the ocean, and, unfortunately, continues over the edge and lands in the ocean. a) Find the car's position relative to the base of the cliff when the car lands in the ocean. b) Find the length of time the car is in the air.

The given variables are: - The angle of the incline: \(\theta = 17^{\circ}\) below horizontal - The distance it rolls down the incline: \(d = 29\mathrm{~m}\) - The height of the cliff: \(h = 55\mathrm{~m}\) - The acceleration due to gravity: \(g = 9.81 \mathrm{m/s^2}\)

Using the angle of the incline, we can find the acceleration of the car down the incline caused by gravity: $$a_{incline} = g \sin \theta = 9.81 \mathrm{m/s^2} \cdot \sin 17^{\circ} = 2.86 \mathrm{m/s^2}$$

We need to find the car's velocity when it reaches the edge of the cliff using the equation: $$v^2 = v_0^2 + 2a_{incline}d$$ Since the initial velocity \(v_0 = 0\), we simplify it to: $$v^2 = 2a_{incline}d$$ Then, we find the magnitude of the velocity at the edge of the cliff: $$v = \sqrt{2 \cdot 2.86 \mathrm{m/s^2} \cdot 29\mathrm{~m}} = 10.37\mathrm{~m/s}$$

Now, we need to find the horizontal and vertical components of the velocity when the car is in the air. The horizontal component of the velocity will be constant because there is no horizontal force acting on the car. To find the horizontal component, we use: $$v_x = v \cos \theta = 10.37\mathrm{~m/s} \cdot \cos 17^{\circ} = 9.86\mathrm{~m/s}$$ Next, we need to find the vertical component of the velocity by: $$v_y = v \sin \theta = 10.37\mathrm{~m/s} \cdot \sin 17^{\circ} = 3.06\mathrm{~m/s}$$

We use the vertical component of the velocity to find the time the car is in the air because the car will fall freely under gravity. We can use the following equation for the vertical motion: $$h = v_y t - \frac{1}{2}gt^2$$ Rearrange to solve for t: $$t = \frac{2(h - v_y t)}{g}$$ Using the quadratic formula: $$t = \frac{-v_y \pm \sqrt{v_y^2 + 2gh}}{g}$$ Selecting the positive value for t, since the negative value does not make physical sense: $$t = \frac{-3.06 \mathrm{~m/s} + \sqrt{(3.06 \mathrm{~m/s})^2 + 2(9.81 \mathrm{~m/s^2})(55 \mathrm{~m})}}{9.81 \mathrm{~m/s^2}} = 3.38\mathrm{~s}$$ Now, we need to find the horizontal distance the car travels while in the air: $$x = v_x t = 9.86 \mathrm{~m/s} \times 3.38\mathrm{~s} = 33.33\mathrm{~m}$$

To find the car's position relative to the base of the cliff, we need to find the horizontal distance that the car traveled on the incline before going over the edge. Using the Pythagorean theorem: $$x_{incline} = d \cos \theta = 29\mathrm{~m} \cdot \cos 17^{\circ} = 27.43\mathrm{~m}$$ Finally, finding the car's position relative to the base: $$x_{relative} = x - x_{incline} = 33.33\mathrm{~m} - 27.43\mathrm{~m} = 5.90\mathrm{~m}$$ a) The car's position relative to the base of the cliff when it lands in the ocean is \(5.90\mathrm{~m}\). b) The length of time the car is in the air is \(3.38\mathrm{~s}\).