Question

A toy rocket is launched with an initial velocity of 12.0 m/s in the horizontal direction from the roof of a 30.0-m-tall building. The rocket's engine produces a horizontal acceleration of \((1.60 m/s^3)t\), in the same direction as the initial velocity, but in the vertical direction the acceleration is \(g\), downward. Ignore air resistance. What horizontal distance does the rocket travel before reaching the ground?

Short answer

The rocket travels approximately 29.71 meters horizontally before hitting the ground.

Step-by-step solution

Analyze the Problem

We're asked to find the horizontal distance traveled by the rocket. Since the rocket is influenced by different accelerations in horizontal and vertical directions, we will solve each independently and then relate them.

Set up Horizontal Motion Equation

The horizontal distance traveled is affected by the initial horizontal velocity and horizontal acceleration. The equation for horizontal displacement is \( x = v_0t + \frac{1}{2}a_xt^2 \). Here, \(v_0 = 12.0\, m/s\) and \(a_x = 1.60 t\, m/s^3\). To find the position function \(x(t)\), we need to integrate the acceleration term.

Integrate Horizontal Acceleration

From the problem, the horizontal acceleration is \( a_x = 1.60t\; m/s^3 \). Integrating this with respect to time gives the velocity function: \( v(t) = \int 1.60t \, dt = 0.80t^2 + C \). At \( t = 0 \), \( v(0) = 12.0 \), so \( C = 12.0 \). Thus, \( v(t) = 0.80t^2 + 12.0 \). Integrating again gives \( x(t) = \int (0.80t^2 + 12) \, dt = \frac{0.80}{3}t^3 + 12t = 0.27t^3 + 12t \).

Set up Vertical Motion Equation

The vertical motion is influenced only by gravity \( g = 9.81\, m/s^2 \) and initial height. The equation for vertical displacement is \( y(t) = y_0 + v_{0y}t - \frac{1}{2}gt^2 \). Given \(y_0 = 30.0\, m\) and \(v_{0y} = 0\), this becomes \( y(t) = 30 - 4.905t^2 \).

Find Time to Hit the Ground

The rocket hits the ground when \( y(t) = 0 \). Solve \( 30 - 4.905t^2 = 0 \) to find \( t \). This simplifies to \( 4.905t^2 = 30 \), resulting in \( t = \sqrt{30/4.905} \approx 2.47\, s \).

Calculate Horizontal Distance

Substitute \( t = 2.47 \) into the horizontal position function: \( x(t) = 0.27(2.47)^3 + 12(2.47) \approx 29.71 \, m \). Thus, the horizontal distance traveled is approximately 29.71 meters.

Key concepts

Horizontal Acceleration

In projectile motion, horizontal acceleration refers to any additional force acting horizontally on an object after it has been launched. Typically, in many projectile problems, the horizontal acceleration is zero unless stated otherwise. This means the object moves horizontally with a constant velocity. However, in our problem, the rocket experiences a unique horizontal acceleration given by \( (1.60 \, \text{m/s}^3)t \). This indicates that the horizontal acceleration increases linearly with time.

To find the horizontal displacement of the rocket, we must first understand this changing acceleration. We start by integrating the acceleration term to obtain the velocity. From the problem, the initial velocity, \( v_0 \), is 12.0 m/s. Integrating the time-dependent acceleration function, \( a_x = 1.60t \), provides us with the velocity function \( v(t) = 0.80t^2 + 12 \). Further integrating gives us the position function, \( x(t) = \frac{0.80}{3}t^3 + 12t \), which we use to determine how far the rocket travels horizontally before hitting the ground.

Understanding how to work with changing accelerations in projectile motion will greatly enhance your problem-solving skills in physics. It allows you to tackle more complex situations beyond simple constant velocity scenarios.

Vertical Motion

Vertical motion in projectile motion is primarily influenced by gravity. This force pulls the object downward, and unless otherwise stated, no other vertical forces act upon it. In the absence of air resistance, which is often assumed in these exercises, vertical motion is determined solely by gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \).

For our toy rocket, we know it is launched from a height of 30 meters. Since initially, there is no vertical velocity component (\( v_{0y} = 0 \)), the vertical displacement can be expressed with the formula \( y(t) = y_0 - \frac{1}{2}gt^2 \), where \( y_0 \) is the starting height. Plugging in the values, we derive the function \( y(t) = 30 - 4.905t^2 \).

To determine when the rocket reaches the ground, solve \( y(t) = 0 \). This involves finding the root of the equation \( 30 - 4.905t^2 = 0 \), solving for \( t \). The solution gives the time \( t \approx 2.47 \) seconds, which is critical for further computations related to horizontal motion. Vertical motion equations ultimately dictate the time constraint for how long the projectile remains in motion.

Trajectory Analysis

Trajectory analysis involves studying the path taken by a projectile as it moves under the influence of various forces. The trajectory of an object in projectile motion is typically parabolic when considering ideal conditions excluding air resistance.

In our exercise with the toy rocket, analyzing the trajectory involved combining the independent influences of horizontal and vertical motion to determine the rocket's path and length of travel. By solving the vertical motion equation first to find the time it takes the rocket to hit the ground, we were able to plug this time into our horizontal movement equation to find the horizontal distance. This approach highlights how interconnected these analyses are.

By understanding the distinct roles of horizontal and vertical influences on trajectory, you're better equipped to predict and calculate various points and distances in the path of a projectile. Mastering trajectory analysis enables you to solve complex physics problems involving motion.