Problem 67 - University Physics with Modern Physics

Question

A cart carrying a vertical missile launcher moves horizontally at a constant velocity of 30.0 m/s to the right. It launches a rocket vertically upward. The missile has an initial vertical velocity of 40.0 m/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

(a) 81.6 m; (b) 244.8 m; (c) Rocket lands directly above the launch point relative to the cart.

Step 1: Identify the problem for Part (a)

To find the maximum height reached by the rocket, we evaluate the vertical motion. The rocket is in free-fall with an initial vertical velocity of 40.0 m/s and an acceleration due to gravity of \( g = -9.8 \text{ m/s}^2 \). The maximum height is reached when the vertical velocity becomes 0 m/s.

Step 2: Calculate the time to reach maximum height

We use the kinematic equation for vertical motion: \( v = u + at \),where \( v = 0 \), \( u = 40.0 \text{ m/s} \), and \( a = -9.8 \text{ m/s}^2 \). Solving for \( t \), \[ 0 = 40.0 \text{ m/s} - 9.8 \text{ m/s}^2 \times t \]\( t = \frac{40.0 \text{ m/s}}{9.8 \text{ m/s}^2} \approx 4.08 \text{ s} \).

Step 3: Calculate maximum height

Using the time to reach maximum height, calculate the maximum height with the formula:\( h = ut + \frac{1}{2}at^2 \),substituting in the values:\[ h = 40.0 \text{ m/s} \times 4.08 \text{ s} + \frac{1}{2} \times (-9.8 \text{ m/s}^2) \times (4.08 \text{ s})^2 \approx 81.6 \text{ m} \].

Step 4: Calculate total time in air

The total time in the air is double the time to reach the maximum height since the ascent time equals the descent time. Thus, \( t_{total} = 2 \times 4.08 \text{ s} \approx 8.16 \text{ s} \).

Step 5: Determine distance cart travels (Part b)

The cart travels horizontally at 30.0 m/s for the duration the rocket is in the air. Using the total time:\( d = v \times t_{total} \ = 30.0 \text{ m/s} \times 8.16 \text{ s} = 244.8 \text{ m} \).

Step 6: Relative landing position of the rocket (Part c)

The horizontal component of the rocket's velocity equals the cart's constant speed (30.0 m/s). Consequently, the rocket moves horizontally at the same speed as the cart throughout its journey. Thus, the rocket lands directly above where it was launched, relative to the cart.

Key Concepts

Kinematic Equations

Kinematic equations are essential tools for solving problems in physics related to motion. These equations describe the relationships between velocity, acceleration, displacement, and time. They are particularly useful in projectile motion, where objects move under the influence of gravity.

In our problem, the kinematic equations help us determine how the rocket travels vertically. The specific equation used here is: \[ v = u + at \] where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

This equation allowed us to find the time it takes for the rocket to reach its maximum height by setting the final velocity (\( v \) ) to zero because at the highest point of vertical motion, the upward speed is zero before it falls back down.

Vertical Motion

Vertical motion focuses on how the rocket moves up and down under the influence of gravity. When discussing vertical motion, it's important to remember that gravity is the only force affecting the rocket's movement after it's launched. This is crucial in determining both the rocket's maximum height and the time it takes to reach it.

The initial vertical velocity given (40.0 m/s) and the constant acceleration due to gravity (\(-9.8 \text{ m/s}^2\)) help us map the rocket's journey in the vertical direction. Using the kinematic equation, we calculated how long it takes the rocket to stop ascending (4.08 seconds).

Once it reaches that point, the rocket begins to descend, and the time it takes to go down is the same, resulting in a total flight time of 8.16 seconds. This information is pivotal when calculating the maximum height reached (81.6 meters), using the equation: \[ h = ut + \frac{1}{2}at^2 \] which considers both the initial velocity and the effect of gravity on the rocket's ascent.

Constant Velocity

Constant velocity plays a key role in the horizontal motion of the cart. While the rocket moves vertically, the cart continues to travel horizontally at a uniform speed of 30.0 m/s.

In physics, constant velocity implies that the speed and direction of an object remain unchanged over time. This characteristic simplifies calculations since the horizontal distance can be found by simply multiplying the velocity by the total time the rocket is in the air.

Thus, the distance travelled by the cart during the rocket's flight is:\[ d = v \times t_{total} = 30.0 \text{ m/s} \times 8.16 \text{ s} = 244.8 \text{ m} \]This shows that while the rocket is in vertical motion, the cart keeps moving steadily along its path.