Problem 13 - College Physics

Question

\(\bullet\) Leaping the river, A \(10,000 \mathrm{N}\) car comes to a bridge during a storm and finds the bridge washed out. The 650 \(\mathrm{N}\) driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 21.3 \(\mathrm{m}\) above the river, while the opposite side is a mere 1.80 \(\mathrm{m}\) above the river. The river itself is a raging torrent 61.0 \(\mathrm{m}\) wide. (a) How fast should the car be traveling just as it leaves the cliff in order to clear the river and land safely on the opposite side? (b) What is the speed of the car just before it lands safely on the other side?

Step-by-Step Solution

Answer

(a) 21.63 m/s; (b) 35.11 m/s.

Step 1: Identify the known values

Given:- The height of the departing cliff: 21.3 m- The height of the opposite side: 1.80 m- The width of the river: 61.0 m- Weight of the car: 10,000 N- Weight of the driver: 650 NGravity, \( g = 9.81 \ \mathrm{m/s^2} \)

Step 2: Calculate the total mass of the car and driver

The total weight is the sum of the car's and driver's weights: \( 10,000 \ \mathrm{N} + 650 \ \mathrm{N} = 10,650 \ \mathrm{N} \).The total mass is calculated by:\[ \text{Mass} = \frac{\text{Total Weight}}{g} = \frac{10,650}{9.81} \approx 1,086.65 \ \mathrm{kg} \]

Step 3: Determine the time of flight

The vertical displacement \( \Delta y \) is given by:\[ \Delta y = - 21.3 \ \mathrm{m} + 1.8 \ \mathrm{m} = -19.5 \ \mathrm{m} \] Using the kinematic equation for vertical motion:\(- \Delta y = v_{0y} t + \frac{1}{2} g t^2\), with initial vertical speed \(v_{0y} = 0\):\[ 19.5 = \frac{1}{2} \times 9.81 \times t^2 \]\[ t = \sqrt{\frac{39}{9.81}} \approx 2.82 \ \mathrm{s} \]

Step 4: Calculate horizontal velocity needed to clear the river

The horizontal distance to cover is 61.0 m. Use the horizontal motion equation:\[ d = v_{0x} \times t \]\[ 61.0 = v_{0x} \times 2.82 \]Solving for \( v_{0x} \):\[ v_{0x} = \frac{61.0}{2.82} \approx 21.63 \ \mathrm{m/s} \]

Step 5: Calculate speed of the car just before landing

Calculate the final vertical velocity \( v_{fy} \) using:\[ v_{fy} = v_{0y} + g \times t = 0 + 9.81 \times 2.82 = 27.65 \ \mathrm{m/s} \]Find the magnitude of the resultant velocity combining horizontal and vertical components:\[ v_{f} = \sqrt{v_{0x}^2 + v_{fy}^2} = \sqrt{21.63^2 + 27.65^2} \]\[ v_{f} \approx \sqrt{467.56 + 764.72} \approx \sqrt{1232.28} \approx 35.11 \ \mathrm{m/s} \]

Key Concepts

Kinematic Equations

Understanding projectile motion starts with the kinematic equations. These equations connect key variables, allowing us to solve complex motion problems by breaking them down into components.

• The primary kinematic equations account for factors like initial velocity, acceleration, time, and displacement. They're essential for predicting an object's motion.
• In vertical motion, the equation \( y = v_{0y} t + \frac{1}{2} g t^2 \) helps us find vertical displacement, where \( y \) is the displacement, \( v_{0y} \) the initial vertical velocity (often zero in free-fall scenarios), \( g \) the acceleration due to gravity \( (9.81 \ \mathrm{m/s^2}) \), and \( t \) is the time.
• Horizontal motion is simpler since it's affected by initial speed and distance without acceleration. \( x = v_{0x} t \) illustrates how horizontal distance \( x \) is covered at a constant velocity.

Breaking problems into these equations allows for precise calculations and a clearer understanding of projectile motion.

Horizontal and Vertical Motion

Projectile motion is a combination of two independent components: horizontal and vertical motion. Understanding the distinction and interaction between these allows for the analysis of complex scenarios like our car leaping across the river.

• Horizontal Motion: - It is freely moving without acceleration (ignoring air resistance) as there is no force acting along the horizontal axis post-launch.
  - The horizontal velocity \( v_{0x} \) remains constant. In our problem, it's determined using the needed velocity to cover the horizontal distance, as expressed in the equation \( d = v_{0x} \times t \).
• Vertical Motion: - Governed by gravity, it follows a deceleration-acceleration path resulting in a parabolic trajectory.
  - The initial vertical component of velocity is often zero in a projectile launched horizontally. Gravity influences it by increasing the downward velocity as per the equation \( v_{fy} = v_{0y} + g \times t \).

This bifurcation ensures distinct analysis paths: predicting how far and how safely the vehicle clears the gap.

Physics Problem Solving

Solving physics problems is about methodically applying concepts to reach a solution. With projectile motion problems, systematic problem-solving makes the calculation of the best launch speed essential.

• **Identify known and unknown variables**: Note what's provided, like heights, widths, and forces, to calculate needed variables like mass or velocity.
• **Break down into components**: - Vertical and horizontal motions are analyzed separately to determine factors like time of flight and distance covered.
  - Formulas specific to each motion aspect (e.g., kinematic equations) are applied.
• **Apply logical reasoning:** - Use reasoning to check if calculated values make sense, given physical constraints.
  - Ensure that results match expected physical behaviors, ensuring the projectile lands correctly as predicted.

This structured approach in physics problem-solving can make even daunting problems approachable, ensuring we arrive at a reliable solution.