You are planning a stunt to be used in an ice skating show. For this stunt a skater will skate down a frictionless ice ramp that is inclined at an angle of \(15.0^{\circ}\) above the horizontal. At the bottom of the ramp, there is a short horizontal section that ends in an abrupt drop off. The skater is supposed to start from rest somewhere on the ramp, then skate off the horizontal section and fly through the air a horizontal distance of $7.00 \mathrm{m}\( while falling vertically for \)3.00 \mathrm{m},$ before landing smoothly on the ice. How far up the ramp should the skater start this stunt?

Using the equation of motion to find the final vertical velocity (\(v_y\)) in terms of the vertical distance fallen (\(-3.00 \mathrm{m}\), taking up as positive), we have: $$v_y^2 = u_y^2 + 2as$$ where - \(v_y\) is the final vertical velocity - \(u_y\) is the initial vertical velocity (0) - \(a\) is the gravitational acceleration (\(9.8 \mathrm{m/s^2}\), taking down as positive) - \(s\) is the vertical distance fallen (\(-3.00 \mathrm{m}\)) Plugging in the values, we get: $$v_y^2 = 0^2 + 2(9.8)(-3) = -58.8$$ Now, finding the square root of both sides: $$v_y = \pm\sqrt{-58.8} = \pm\sqrt{58.8\mathrm{i}^2} = \pm7.67\mathrm{i}$$ Since \(v_y\) is downward, we choose the negative value: $$v_y = -7.67 \mathrm{m/s}$$

Using the horizontal distance traveled (\(7.00 \mathrm{m}\)), and the relationship between the total time of flight (\(t_{total}\)) and the vertical distance fallen: $$t_{total} = \frac{2s}{v_y}$$ Now, we have: $$t_{total} = \frac{2(-3.00)}{-7.67} \approx 0.784 \mathrm{s}$$ We know that the horizontal velocity (\(v_x\)) is constant. So, we can find it using this relationship: $$v_x = \frac{d_x}{t}$$ where: - \(d_x\) is the horizontal distance traveled (\(7.00 \mathrm{m}\)) - \(t\) is the total time of flight Now, we have: $$v_x = \frac{7.00}{0.784} \approx 8.93 \mathrm{m/s}$$

Using the relationship between the components of the velocity and the initial velocity on the ramp (\(v\)), we have: $$v^2 = v_x^2 + v_y^2$$ Plugging in the values, we get: $$v^2 = (8.93)^2 + (-7.67)^2 \approx 140.26$$ Now, finding the square root: $$v = \sqrt{140.26} \approx 11.8 \mathrm{m/s}$$

Using the equation of motion to find the distance up the ramp (\(d\)), and considering the acceleration on the ramp (\(a_r = -9.8\sin{15^\circ}\)) and the angle of the ramp (\(15.0^\circ\)), we have: $$v^2 = u^2 + 2a_rd$$ which can be rearranged as: $$d = \frac{v^2 - u^2}{2a_r}$$ Plugging in the values, we get: $$d = \frac{(11.8)^2 - 0^2}{2(-9.8\sin{15^\circ})} \approx 16.3 \mathrm{m}$$ So, the skater should start the stunt at the distance of approximately 16.3 meters up the ramp.