Problem 49 - College Physics

Question

\bullet A cart carrying a vertical missile launcher moves horizontally at a constant velocity of 30.0 \(\mathrm{m} / \mathrm{s}\) to the right. It launches a rocket vertically upward. The missile has an initial vertical velocity of 40.0 \(\mathrm{m} / \mathrm{s}\) relative to the cart. (a) How high does the rocket go? (b) How far does the cart travel while the rocket is in the air? (c) Where does the rocket land relative to the cart?

Step-by-Step Solution

Answer

Maximum height of the rocket is 81.6 meters. The cart travels 244.8 meters while the rocket is in the air. The rocket lands at the same position relative to the cart.

Step 1: Determine Time of Flight

To determine how high the rocket goes and where it will land, we first need to determine the time of flight. The rocket is only influenced by gravity after launch.We use the formula for vertical motion: \[ v = u + at \]where \(v\) is the final velocity (0 m/s at the peak), \(u\) is the initial vertical velocity (40.0 m/s), \(a\) is the acceleration due to gravity (-9.8 m/s²).Solving for time \(t\) when \(v = 0\):\[ 0 = 40.0 - 9.8t \]\[ t = \frac{40.0}{9.8} \approx 4.08 \text{ s} \]This is the time to reach the peak, so the total time of flight (up and down) is \(2 \times 4.08 = 8.16\) seconds.

Step 2: Calculate Maximum Height

Using the time to reach the maximum height (4.08 seconds), we calculate how high the rocket goes using:\[ s = ut + \frac{1}{2} a t^2 \]where \(s\) is the maximum height, \(u\) is the initial velocity (40.0 m/s), \(a\) is the acceleration due to gravity (-9.8 m/s²), and \(t\) is the time to reach the maximum height (4.08 s):\[ s = 40.0 \times 4.08 + \frac{1}{2} \times (-9.8) \times (4.08)^2 \]\[ s = 163.2 - 81.6 \]\[ s = 81.6 \text{ meters} \]

Step 3: Calculate Horizontal Distance Traveled by Cart

The cart moves at a constant horizontal velocity of 30.0 m/s. To find the distance the cart travels while the rocket is in the air (8.16 seconds), we use the formula:\[ d = v \times t \]where \(d\) is the horizontal distance, \(v\) is the velocity of the cart (30.0 m/s), and \(t\) is the total time the rocket is in the air (8.16 s):\[ d = 30.0 \times 8.16 \]\[ d = 244.8 \text{ meters} \]

Step 4: Determine Rocket's Landing Point Relative to the Cart

Since the horizontal velocity of the rocket is the same as the cart's (30.0 m/s) due to the relative motion, the rocket continues to move horizontally alongside the cart throughout its flight. Therefore, the rocket will land exactly at the same horizontal position as the cart. Thus, the rocket lands at the same point relative to the cart.

Key Concepts

Constant Velocity

In physics, when an object moves at a constant velocity, it means that both its speed and direction remain unchanged over time. For our exercise, the cart moves horizontally with a constant velocity of 30.0 m/s to the right.

This implies that there is no acceleration acting in the horizontal direction. The horizontal motion of the cart can thus be described by the basic formula:

• \( d = v \times t \)

where \(d\) represents the distance traveled, \(v\) is the constant velocity, and \(t\) is the time.

Since there is no change in velocity, this simplifies the calculations we need to perform for the horizontal motion, enabling us to focus mainly on the vertical motion of the rocket.

Vertical Launch

A vertical launch in projectile motions refers to the scenario where an object is projected straight up into the air. In the exercise, our missile launcher propels the rocket upwards at an initial velocity of 40.0 m/s.

The vertical motion is governed by the principles of kinematics, and notably affected by gravity, which constantly acts downward. Thus, while calculating the rocket's maximum height or time of flight, only vertical velocity components and gravitational acceleration are considered.

The initial vertical launch velocity helps determine important metrics like:

• The maximum height using \(s = ut + \frac{1}{2} a t^2\).
• The time required to reach the peak height, where the upward velocity becomes zero.

This highlights the decoupling of vertical and horizontal motions, central in solving projectile problems.

Time of Flight

Time of flight is a crucial concept in projectile motion, representing the total duration the object spends in the air, from the moment of launch until it returns to the same vertical level.

In our specific exercise, we determine the time it takes for the rocket to first reach its peak height, then double it to include the descending journey. The formula used is:

• \( v = u + at \)

where \(v\) is the final velocity at the peak (0 m/s), leading to the calculation.Once the peak's time is determined to be approximately 4.08 seconds, the total flight time becomes 8.16 seconds by multiplying the ascent time by two.

This entire duration considers the effect of gravity, with no external forces acting after the initial launch. Understanding this helps in synchronizing the concurrent horizontal and vertical motions.

Acceleration Due to Gravity

Acceleration due to gravity is a constant force acting on projectiles, pulling them towards the Earth. Denoted by \(-9.8 \text{ m/s}^2\), it influences only the vertical motion of projectiles.

After the rocket's launch, gravity ensures that the initial vertical velocity decreases steadily until the peak is reached, where the velocity becomes zero. This point marks the highest point of the flight.

• During rise: Velocity decreases by 9.8 m/s every second.
• During descent: Velocity increases by 9.8 m/s every second back down.

Using gravity, we calculate maximum height and time of flight. Importantly, gravity acts independently of horizontal motion, simplifying calculations by allowing us to address vertical and horizontal components separately.