91Ó°ÊÓ

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.