Problem 62 - College Physics

Question

\cdotse: Leaping the river, II. A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle. (See Figure \(3.45 .\) ) The takeoff ramp was inclined at \(53.0^{\circ},\) the river was 40.0 \(\mathrm{m}\) wide, and the far bank was 15.0 \(\mathrm{m}\) lower than the top of the ramp. The river itself was 100 \(\mathrm{m}\) below the ramp. You can ignore air resistance. (a) What should his speed have been at the top of the ramp for him to have just made it to the edge of the far bank? (b) If his speed was only half the value found in \((a)\) where did he land?

Step-by-Step Solution

Answer

(a) Calculate initial speed using quadratic motion equations. (b) New landing distance with half speed.

Step 1: Analyze the problem

We need to find the initial speed required to jump a distance (horizontal range) and land at a different elevation. We'll look at both horizontal and vertical motion components.

Step 2: Break down initial velocity into components

The ramp is inclined at \(53.0^{\circ}\). Use trigonometry to find the horizontal \((v_{x})\) and vertical \((v_{y})\) components of the initial velocity \(v_{0}\). \[v_{x} = v_{0} \cos(53.0^{\circ})\]\[v_{y} = v_{0} \sin(53.0^{\circ})\]

Step 3: Determine time for horizontal motion

The horizontal distance to be covered is \(40.0 \ m\). The time \(t\) for this horizontal motion can be given by \[t = \frac{\text{Horizontal Distance}}{v_{x}} = \frac{40.0}{v_{0} \cos(53.0^{\circ})}\]

Step 4: Account for vertical motion

The vertical motion needs to account for a drop of \(15.0 \ m\). The equation of motion in vertical direction is given by:\[\Delta y = v_{y} t - \frac{1}{2}gt^2\]Substitute \(\Delta y = -15.0\), and solve for \(t\) using the expression for \(v_{y}\) and \(v_{x}\).

Step 5: Solve for initial speed

From Steps 3 and 4, equate the two expressions for \(t\) and solve for \(v_{0}\). This will result in a quadratic equation in terms of \(v_{0}\). Solve this quadratic to find the initial speed needed.

Step 6: Calculate modified landing position

If his speed is half the calculated \(v_{0}\), use the horizontal motion equation to find the new horizontal distance. Use: \[v_{0}' = \frac{v_{0}}{2}\]\[\text{New Distance} = v_{0}' \cos(53.0^{\circ}) \times t\]where \(t\) is determined from vertical motion.

Key Concepts

Inclined Plane

Projectile motion often involves an inclined plane, which is simply a flat surface tilted at an angle to the horizontal. Here, the ramp is the inclined plane, tilted at an angle of 53 degrees. Understanding this concept is crucial because it informs how we derive the components of the initial velocity using basic trigonometry.

Inclined planes are a fundamental topic in physics. They allow us to break a force or velocity into two components: horizontal and vertical. In this problem, the motorcycle's velocity as it leaves the ramp is divided into a horizontal and a vertical component.

This division helps us solve the problem by analyzing each component separately using the laws of physics.

Kinematic Equations

Kinematic equations describe the motion of objects in terms of displacement, velocity, and acceleration. These equations allow us to solve for one variable if the others are known. Here, they help determine the time of flight and the required initial velocity.

For vertical motion, we use the equation of motion:

• y = v_{y} t - \frac{1}{2}gt^2

where \(v_{y}\) is the vertical component of the initial velocity, \(g\) is the acceleration due to gravity, and \(t\) is the time. This helps determine how long the projectile (motorcycle) is in the air.

For horizontal motion, the equation is simpler as there is no horizontal acceleration:

• x = v_{x} t

Here, \(v_{x}\) is the horizontal component of the initial velocity. By solving these equations, we can find the initial speed needed to just reach the far bank.

Physics Problem Solving

Solving physics problems like this one can seem daunting at first, but following a structured approach can simplify the task.

First, understand the problem and identify what you need to find. Next, break down the motion into components and use appropriate equations for each.

In this case, we identified that the problem involves motion over an inclined plane and a change in vertical position. We then used trigonometry to determine velocity components and applied kinematic equations to solve for time and distance.

By solving step-by-step and ensuring each part of the problem is addressed, such problems become more manageable. It’s about translating the real-world scenario into mathematical equations, solving those equations, and then interpreting the solution back in terms of the original problem.

Trigonometry in Physics

Trigonometry plays a critical role in physics, especially when dealing with inclined planes and projectile motion. It helps in breaking down a vector into its components, which simplifies the problem.

In physics problems like this one, trigonometry is used to separate the initial velocity into its horizontal and vertical components.

• \(v_x = v_0 \cos(\theta)\)
• \(v_y = v_0 \sin(\theta)\)

where \(v_{0}\) is the initial velocity and \(\theta\) is the angle of the incline. This decomposition is essential because it allows the use of kinematic equations for solving two-dimensional motion problems.

Even though the trigonometric identities involved are basic, they are vital in a wide array of physics applications, from solving simple projectile motion to more complex scenarios.