A Motorcycle Jump. You are planning to make a jump with your motorcycle by driving over a ramp that will launch you at an angle of 30.0∘ with respect to the horizontal. The front edge of the ramp on which you are supposed to land, however, is 25.0 ft lower than the edge of the launch ramp (i.e., your launch height).
(a) Assuming a launch speed of 70.0 mph, at what horizontal distance from your launch point should the landing ramp be placed?
(b) In order to land smoothly, the angle of the landing ramp should match the direction of your velocity vector when you touch down. What should be the angle of the landing ramp?
To solve this problem, we will first convert the launch speed from mph to ft/s:
\(70.0 \text{ mph} = \frac{70.0 \times 5280}{3600} \text{ ft/s} = 102.67 \text{ ft/s}\)
(a) To find the horizontal distance from the launch point to the landing point, we can use the kinematic equation:
\(y = x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta}\)
where:
\(y = 25.0 \text{ ft}\) (vertical distance between launch and landing points)
\(\theta = 30.0^\circ\)
\(g = 32 \text{ ft/s}^2\) (acceleration due to gravity)
\(v_0 = 102.67 \text{ ft/s}\)
Substitute the given values and solve for x:
\(25.0 = x\tan\left(30.0^\circ\right) - \frac{32x^2}{2(102.67)^2\cos^2\left(30.0^\circ\right)}\)
\(25.0 = x\left(0.577\right) - \frac{32x^2}{2(105.72)^2(0.75)}\)
\(25.0 = 0.577x - 0.0891x^2\)
Solving the quadratic equation, we get two possible solutions for x:
\(x = 136.9 \text{ ft}\) and \(x = 228.0 \text{ ft}\)
Therefore, the landing ramp should be placed at a horizontal distance of 136.9 ft or 228.0 ft from the launch point for a smooth landing.
(b) To find the angle of the landing ramp, we need to find the angle at which the velocity vector will be at touch down. This can be done by finding the angle between the horizontal and the velocity vector when landing:
\(\tan\phi = \frac{v_y}{v_x}\)
where:
\(v_y = v_0\sin\theta\) (vertical component of velocity)
\(v_x = v_0\cos\theta\) (horizontal component of velocity)
\(v_0 = 102.67 \text{ ft/s}\)
\(\theta = 30.0^\circ\)
Substitute the given values:
\(\tan\phi = \frac{102.67(0.5)}{102.67(0.866)}\)
\(\tan\phi = \frac{51.34}{89.02}\)
\(\tan\phi = 0.5768\)
\(\phi = \tan^{-1}(0.5768)\)
\(\phi = 29.4^\circ\)
Therefore, the angle of the landing ramp should be approximately 29.4 degrees in order to match the direction of the velocity vector at touch down.