Physics of a Bicycle Jump: Flying Over a Container
Analyzing the physics behind a bicycle's jump over a container using kinematics and projectile motion.
Problem Statement
Imagine a daredevil cyclist attempting to jump a container from the roof of a building. The building has a height of $H = 5.0 \, \text{m}$, and the container placed adjacent to the building has dimensions $h = 3.0 \, \text{m}$ (height) and $w = 4.5 \, \text{m}$ (width). The cyclist and bicycle are modeled as a point mass, and we want to answer the following questions:
- What is the minimum initial velocity $v_0$ required for the cyclist to clear the container without hitting it?
- If the cyclist jumps with the minimum initial velocity, what is the total horizontal distance $L$ they cover before landing back on the ground?
- What is the angle $\alpha$ of the bicycle’s velocity with respect to the horizontal at the moment of landing?
Assumptions and Setup
- The reference point (origin) is at the launch point.
- Positive $x$ points right (horizontal direction), and positive $y$ points downward (vertical direction).
- Gravity acts downward with acceleration $g = 10 \, \text{m/s}^2$.
- The trajectory of the bicycle follows the equations of projectile motion.
(a) Minimum Initial Velocity $v_0$
The bicycle must fly over the container without hitting it. This means the trajectory of the bicycle must pass through the top-right corner of the container at point $(w, H-h)$, where $w = 4.5 \, \text{m}$ and $H - h = 2.0 \, \text{m}$.
Vertical Motion Equation:
The vertical position $y$ of the bicycle is given by:
\[y = \frac{1}{2} g t^2\]Horizontal Motion Equation:
The horizontal position $x$ of the bicycle is given by:
\[x = v_0 t \quad \Rightarrow \quad t = \frac{x}{v_0}\]Substituting $t = \frac{x}{v_0}$ into the vertical motion equation:
\[y = \frac{1}{2} g \left(\frac{x}{v_0}\right)^2 \quad \Rightarrow \quad y = \frac{1}{2} \frac{g}{v_0^2} x^2\]At the point $(w, H-h)$:
\[H - h = \frac{1}{2} \frac{g}{v_0^2} w^2\]Rearranging for $v_0$:
\[v_0 = \sqrt{\frac{\frac{1}{2} g w^2}{H - h}}\]Substitute $g = 10 \, \text{m/s}^2$, $w = 4.5 \, \text{m}$, and $H - h = 2.0 \, \text{m}$:
\[v_0 = \sqrt{\frac{\frac{1}{2} \cdot 10 \cdot 4.5^2}{2.0}} \approx 7.12 \, \text{m/s}\]The minimum initial velocity required is:
\[v_0 \approx 7.12 \, \text{m/s}\](b) Total Horizontal Distance $L$
The total horizontal distance $L$ corresponds to the point where the cyclist lands back at the ground ($y = H$). From the vertical motion equation:
\[y = \frac{1}{2} \frac{g}{v_0^2} x^2\]At the landing point $(L, H)$, substitute $y = H$:
\[H = \frac{1}{2} \frac{g}{v_0^2} L^2 \quad \Rightarrow \quad L = \sqrt{\frac{2 H v_0^2}{g}}\]Substitute $v_0^2 = \frac{\frac{1}{2} g w^2}{H - h}$ from part (a):
\[L = \sqrt{\frac{2 H \cdot \frac{\frac{1}{2} g w^2}{H - h}}{g}}\]Simplify:
\[L = \sqrt{\frac{H w^2}{H - h}}\]Substitute $H = 5.0 \, \text{m}$, $w = 4.5 \, \text{m}$, and $H - h = 2.0 \, \text{m}$:
\[L = \sqrt{\frac{5.0 \cdot 4.5^2}{2.0}} \approx 7.12 \, \text{m}\]The total horizontal distance is:
\[L \approx 7.12 \, \text{m}\](c) Angle of Velocity $\alpha$ at Landing
At the landing point, the components of velocity are:
Horizontal Velocity:
The horizontal velocity remains constant throughout the motion:
\[v_{f,x} = v_0\]Vertical Velocity:
From the vertical motion equation:
\[v_{f,y} = g t\]At the landing point, $t = \frac{L}{v_0}$, so:
\[v_{f,y} = g \frac{L}{v_0}\]The angle $\alpha$ of the velocity vector with respect to the horizontal is:
\[\alpha = \arctan\left(\frac{v_{f,y}}{v_{f,x}}\right) = \arctan\left(\frac{g L}{v_0^2}\right)\]Substitute $v_0^2 = \frac{\frac{1}{2} g w^2}{H - h}$ and $L = \sqrt{\frac{H w^2}{H - h}}$:
\[\alpha = \arctan\left(\frac{2 H}{\sqrt{\frac{H w^2}{H - h}}}\right)\]Simplify further and substitute values:
\[\alpha = \arctan\left(\frac{2 \cdot 5.0}{\sqrt{\frac{5.0 \cdot 4.5^2}{2.0}}}\right) \approx 55^\circ\]The angle of velocity at landing is:
\[\alpha \approx 55^\circ\]Conclusion
The analysis reveals the following:
- The minimum initial velocity required to clear the container is approximately $v_0 \approx 7.12 \, \text{m/s}$.
- The total horizontal distance covered is $L \approx 7.12 \, \text{m}$.
- The angle of velocity with the horizontal at landing is $\alpha \approx 55^\circ$.
This problem beautifully illustrates the principles of projectile motion and how physics governs the daring feats of cyclists and stunt performers!