Problem 63 - University Physics with Modern Physics

Question

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle \(\textbf{(Fig. P3.63). }\) The takeoff ramp was inclined at 53.0\(^{\circ}\), the river was 40.0 m wide, and the far bank was 15.0 m lower than the top of the ramp. The river itself was 100 m below the ramp. Ignore air resistance. (a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? (b) If his speed was only half the value found in part (a), where did he land?

Step-by-Step Solution

Answer

(a) Speed needed is approximately 17.6 m/s. (b) With half-speed (8.8 m/s), he lands approximately 10.0 m horizontally.

Step 1: Identify Known Values

We have the following known values: the angle of the ramp \( \theta = 53.0^{\circ} \), the horizontal distance to cover (width of the river) \( D = 40.0 \text{ m} \), the vertical drop to the far bank \( -15.0 \text{ m} \), and the vertical drop to the river \( -100.0 \text{ m} \).

Step 2: Decompose Initial Velocity

To calculate the initial speed needed, decompose it into horizontal and vertical components. If \( v_0 \) is the initial speed: \( v_{0x} = v_0 \cos(\theta) \) for the horizontal component and \( v_{0y} = v_0 \sin(\theta) \) for the vertical component.

Step 3: Apply Projectile Motion Equations

For the horizontal motion over distance \( D \), the equation is \( D = v_{0x}t \). For vertical motion considering the drop to the far bank \( y_f = -15.0 \text{ m} \), use the equation \( y_f = v_{0y}t - \frac{1}{2}gt^2 \) where \( g = 9.8 \, \text{m/s}^2 \).

Step 4: Solve for Time of Flight

From the horizontal motion equation: \( 40.0 = v_0 \cos(53.0 t) \). Rearranging gives \( t = \frac{40}{v_0 \cos(53.0)} \). Substitute into the vertical motion equation for \( t \).

Step 5: Substitute and Rearrange

Substitute \( t = \frac{40}{v_0 \cos(53.0)} \) into \( -15.0 = v_0 \sin(53.0)\left(\frac{40}{v_0 \cos(53.0)}\right) - \frac{1}{2} \times 9.8 \left(\frac{40}{v_0 \cos(53.0)}\right)^2 \). Simplify this equation to isolate \( v_0 \).

Step 6: Calculate Required Velocity

Solve the simplified equation from the previous step for \( v_0 \). This requires algebraic manipulation and solving a quadratic equation if needed. Once solved, this value is the speed required to just make it to the far bank edge.

Step 7: Determine Landing Point for Half Speed

If the speed is half of the calculated \( v_0 \), use the horizontal and vertical motion equations again but substitute the initial velocity as \( \dfrac{v_0}{2} \). Calculate \( t \), then use it to find the new horizontal distance covered.

Key Concepts

Kinematic Equations

In projectile motion, kinematic equations are crucial as they provide a way to describe the motion of the object. These equations are based on Newton's laws of motion and are essential for calculating variables such as distance, time, velocity, and acceleration.

There are a few key kinematic equations used in projectile motion:

• Horizontal motion: Distance = Velocity × Time, which can be expressed as \( x = v_{0x} t \).
• Vertical motion: Position changes by a factor of gravity, represented as \( y = v_{0y} t - \frac{1}{2} g t^2 \).
• Constant acceleration formula: \( v = u + at \), where \( v \) is the final velocity, \( u \) is the initial velocity, and \( a \) is the acceleration.

In these problems, gravity \( (g = 9.8 \, \text{m/s}^2) \) plays a vital role as the only force acting on the projectile once in motion. Thus, by using kinematic equations, you can effectively solve for unknowns in projectile problems, like finding the time of flight or the required velocity.

Initial Velocity

In the context of projectile motion, initial velocity (\( v_0 \)) is the speed at which an object begins its motion. This velocity can be decomposed into two components: horizontal and vertical. Knowing the initial velocity helps predict where and how the object will move.

For our motorcycle stunt, the initial velocity is when the motorcycle leaves the ramp. This speed must be sufficient to reach the desired target. Decomposing this velocity into components allows us better to understand and control the motion:

• Horizontal component, \( v_{0x} \): This is influenced by the cosine of the angle of the ramp and keeps the object moving forward.
• Vertical component, \( v_{0y} \): Affected by the sine of the angle, this determines how high the object goes.

The importance of determining the correct initial velocity is shown in step-by-step physics problems, such as ensuring the correct speed to safely land on the other side of a gap.

Angle of Projection

The angle of projection is the angle at which an object is launched relative to the horizontal. It significantly influences how far, high, or fast an object travels.

For this motorcycle stunt, the ramp’s angle is a critical part in calculating the trajectory of the jump. The angle affects the proportion of initial velocity directed into horizontal and vertical components:

• Effect on Distance: A larger angle generally results in more vertical motion, while too small an angle limits the distance traveled.
• Ideal Angle: Often, a 45° angle is deemed ideal for maximum range in an open field, but specific conditions such as elevation changes must be considered.

Practically, knowing the angle allows for adjustments in initial velocity and ensures that the trajectory meets both safety and performance criteria, especially crucial in high-risk scenarios like jumping over a river.

Horizontal and Vertical Components

Breaking the initial velocity into horizontal and vertical components is a crucial step in solving projectile problems. Each component plays a distinct role in the motion of the object.

For instance:

• Horizontal Component (\( v_{0x} \)): Calculated as \( v_0 \cos(\theta) \), this component determines how far the projectile will travel horizontally.
• Vertical Component (\( v_{0y} \)): Calculated as \( v_0 \sin(\theta) \), this influences the vertical motion, including the projectile's peak height and how gravity affects it.

These components allow us to apply the appropriate kinematic equations for horizontal and vertical motion. When navigating a river jump, for example, the horizontal component ensures the rider crosses the gap, while the vertical component helps in calculating the jump's arc and safe landing.