Problem 70 - University Physics with Modern Physics

Question

A student sits atop a platform a distance \(h\) above the ground. He throws a large firecracker horizontally with a speed \(v\). However, a wind blowing parallel to the ground gives the firecracker a constant horizontal acceleration with magnitude \(a\). As a result, the firecracker reaches the ground directly below the student. Determine the height \(h\) in terms of \(v\), \(a\), and \(g\). Ignore the effect of air resistance on the vertical motion.

Step-by-Step Solution

Answer

The height \(h\) is given by \(h = \frac{2g v^2}{a^2}\).

Step 1: Analyze Horizontal Motion

In the horizontal direction, we'll use the equations of motion to analyze the effects of initial speed and acceleration due to wind. The initial horizontal velocity is \(v\) and the constant horizontal acceleration is \(a\). The time \(t\) taken for the firecracker to reach the ground can be found using: \[ x = vt + \frac{1}{2} a t^2 \]where \(x = 0\) because it directly lands below the starting point.

Step 2: Solve for Time of Flight

Since \(x = 0\), the displacement in horizontal direction is zero. We can set up the equation:\[ 0 = v t + \frac{1}{2} a t^2 \]Simplifying this expression gives:\[ t (v + \frac{1}{2} a t) = 0 \]Since \(t eq 0\), we solve for \(t\): \[ v + \frac{1}{2} a t = 0 \]\[ t = -\frac{2v}{a} \]

Step 3: Analyze Vertical Motion

Vertical motion is influenced only by gravity. We need to find the time it takes for the firecracker to hit the ground. Given that the initial vertical velocity is 0 and the gravitational acceleration is \(g\), the height \(h\) can be calculated using:\[ h = \frac{1}{2} g t^2 \]

Step 4: Substitute Time into Vertical Motion Equation

Substitute the expression for time \(t = -\frac{2v}{a}\) from Step 2 into the vertical motion equation:\[ h = \frac{1}{2} g \left( -\frac{2v}{a} \right)^2 \]Simplify the expression:\[ h = \frac{1}{2} g \left( \frac{4v^2}{a^2} \right) \]\[ h = \frac{2g v^2}{a^2} \]

Key Concepts

Horizontal Motion

When analyzing projectile motion, it's essential to understand horizontal motion. In this scenario, horizontal movement is influenced both by an initial velocity and a constant acceleration due to wind. The path of the firecracker is determined by its initial speed, denoted as \( v \), and the acceleration \( a \), which is parallel to the ground.

The horizontal displacement equation is:

• \( x = vt + \frac{1}{2} a t^2 \)

For this problem, the firecracker lands directly beneath its starting point, making \( x = 0 \). This implies that the sum of the initial velocity multiplied by time and the product of acceleration and time squared (halved) results in no net displacement horizontally. This is crucial for finding the time of flight.

Vertical Motion

Vertical motion in projectile problems is primarily affected by gravitational forces. Here, the firecracker starts with an initial vertical velocity of zero, and gravity is the only force acting vertically on it.

Gravity accelerates the firecracker downwards at a rate of \( g \), the acceleration due to gravity. The core equation for vertical motion, given the starting parameters, is:

• \( h = \frac{1}{2} g t^2 \)

This reveals that the height \( h \) from which the projectile was thrown is related to the time \( t \) it spends in the air and the gravitational pull.

Equations of Motion

Equations of motion are crucial for calculating and understanding the path of projectiles. In this exercise, we apply these equations both horizontally and vertically to discover various elements of motion.

For horizontal motion, where acceleration due to the wind is present, the equation:

• \( x = vt + \frac{1}{2} a t^2 \)

is used to determine the duration of flight.

For vertical motion, the equation:

• \( h = \frac{1}{2} g t^2 \)

links the time of flight to the height from which the projectile begins. Together, these equations help understand the firecracker's full trajectory.

Gravitational Acceleration

Gravitational acceleration, usually denoted by \( g \), is a constant force pulling objects toward the earth's center. This acceleration is approximately \( 9.8 \, \text{m/s}^2 \) on the surface of the Earth and plays a vital role in any type of projectile motion analysis.

In this scenario, since the firecracker is thrown horizontally, gravitational acceleration is the only force acting vertically. It dictates how long the firecracker is in the air by influencing the vertical motion equation:

• \( h = \frac{1}{2} g t^2 \)

Through knowing \( g \), along with time determined from horizontal motion calculations, we can derive how far the firecracker falls, symbolized by the height \( h \). This intertwines gravitational forces with projectile motion to help predict outcomes accurately.