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A cart carrying a vertical missile launcher moves horizontally at a constant velocity of \(30.0 \mathrm{~m} / \mathrm{s}\) to the right (Figure 3.39 ). 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?

Short Answer

Expert verified
(a) The rocket reaches a maximum height of approximately 81.58 m. (b) The cart travels about 244.8 m while the rocket is in the air. (c) The rocket lands at the same position relative to the cart.

Step by step solution

01

Determine the maximum height of the rocket

The rocket's initial vertical velocity is given as \( v_{0y} = 40.0 \ \mathrm{m/s} \). Using the formula for vertical motion, where the final velocity \( v_y \) at the maximum height is 0, and acceleration \( a = -9.81 \ \mathrm{m/s^2} \) (due to gravity), we use the equation: \[ v_y^2 = v_{0y}^2 + 2a y_{max} \] Setting \( v_y = 0 \), we solve for \( y_{max} \): \[ 0 = (40.0)^2 + 2(-9.81) y_{max} \] \[ 0 = 1600 - 19.62 y_{max} \] \[ 19.62 y_{max} = 1600 \] \[ y_{max} = \frac{1600}{19.62} \approx 81.58 \ \mathrm{m} \] Thus, the maximum height reached by the rocket is approximately \( 81.58 \ \mathrm{m} \).
02

Calculate the total time the rocket is in the air

To find the time the rocket takes to reach the maximum height, use the vertical motion equation: \[ v_y = v_{0y} + at \] Set \( v_y = 0 \): \[ 0 = 40.0 - 9.81 t_{up} \] \[ t_{up} = \frac{40.0}{9.81} \approx 4.08 \ \mathrm{s} \] The total time in the air is twice the time to reach maximum height because the upward and downward journeys take equal time when air resistance is negligible. Thus, \[ t_{total} = 2 \times 4.08 = 8.16 \ \mathrm{s} \] The total air time for the rocket is approximately \( 8.16 \ \mathrm{s} \).
03

Calculate the distance traveled by the cart

The cart moves at a constant velocity of \( 30.0 \ \mathrm{m/s} \) horizontally. The total distance traveled by the cart while the rocket is in the air can be calculated using: \[ d = v \times t \] Here, \( v = 30.0 \ \mathrm{m/s} \) and \( t = 8.16 \ \mathrm{s} \): \[ d = 30.0 \times 8.16 \approx 244.8 \ \mathrm{m} \] The cart travels approximately \( 244.8 \ \mathrm{m} \) while the rocket is in the air.
04

Determine where the rocket lands relative to the cart

Since the rocket was launched vertically and lands back vertically with respect to the cart's motion, it experiences the same horizontal velocity as the cart (no horizontal acceleration in the rocket's frame). Hence, the horizontal displacement relative to the cart remains zero. Thus, the point where the rocket lands is directly above where it was launched relative to the cart. The rocket lands at the same horizontal position relative to the cart.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertical Motion
Vertical motion involves the movement of an object in a straight line perpendicular to the horizontal component, primarily influenced by forces such as gravity. In the context of our exercise, a missile is launched vertically from a moving cart. The initial vertical velocity given is \(40 \ \mathrm{m/s}\), which impacts the motion wholly separate from the horizontal motion of the cart.

To calculate how high the rocket travels, we apply the vertical motion formula:
  • Initial vertical velocity \( v_{0y} = 40.0 \ \mathrm{m/s} \)
  • Acceleration due to gravity \( a = -9.81 \ \mathrm{m/s^2} \)
  • Final vertical velocity \( v_y = 0 \), which is the velocity at maximum height
Using the equation \( v_y^2 = v_{0y}^2 + 2ay_{\text{max}} \), we find that the rocket reaches a maximum height of approximately \(81.58 \ \mathrm{m}\). This calculation solely depends on the rocket's vertical launch speed and the constant force of gravity pulling it downward.
Horizontal Motion
Horizontal motion is the motion that happens parallel to the ground. In projectile problems, horizontal motion is often uniform, meaning that it occurs at a constant speed.

For the cart carrying the rocket, its horizontal motion operates independently from the vertical motion of the rocket. The cart moves at \(30.0 \ \mathrm{m/s}\) continuously. As the rocket moves vertically, its horizontal velocity is the same as the cart’s due to it being launched from the cart:
  • Constant horizontal speed \( v_x = 30.0 \ \mathrm{m/s} \)
  • Horizontal distance traveled while the rocket is in the air \( d = v \times t \)

To calculate the distance covered by the cart during the rocket's flight time \( t = 8.16 \ \mathrm{s} \), we have \( d = 30.0 \times 8.16 = 244.8 \ \mathrm{m}\). This simple multiplication reflects how horizontal movement is frequently uniform, marked by a lack of acceleration, assuming no other forces like air resistance.
Relative Velocity
Relative velocity examines how the speed and direction of one object is perceived from another object's perspective. In our scenario, understanding that the rocket maintains the same horizontal speed as the cart is central to grasping the idea of relative velocity.

From the perspective of someone on the cart, the missile's vertical launch and eventual landing exhibit no horizontal movement. However, from a stationary viewer on the ground, the missile appears to move horizontally alongside the cart. This stationary bystander sees the rocket traveling the same horizontal distance as the cart:
  • Rocket's horizontal speed relative to the cart is \(0 \ \mathrm{m/s}\)
  • Rocket's horizontal speed relative to the ground is \(30.0 \ \mathrm{m/s}\)
  • Thus, no net horizontal displacement relative to the cart
Hence, relative velocity helps evaluate where the rocket lands. It underscores that from the cart’s perspective, the rocket comes down in the same vertical position from where it was launched.
Gravity Effect
Gravity is a force that attracts two bodies towards each other, typically pulling objects towards the Earth's center in our day-to-day life. In projectile motion, gravity influences vertical motion by acting as a constant downward force, causing deceleration when moving upwards and acceleration when moving downwards.

In the exercise, gravity is the factor that causes the rocket to decelerate as it rises until it reaches the apex of its trajectory. At this point, vertical velocity is zero. As the rocket descends, gravity accelerates it downward. The key aspects are:
  • Gravitational acceleration is \(a = -9.81 \ \mathrm{m/s^2}\)
  • Gravity does not affect horizontal motion, thus horizontal motion remains constant
  • In the absence of other forces (like significant air resistance), gravity is the sole vertical force

Due to these effects, the total time the rocket spends in the air can be calculated easily through consistent upward and downward travel. Gravity ensures that the rocket's motion is symmetric, where the time taken to ascend equals the time to descend, at \(8.16 \ \mathrm{s}\). This symmetry simplification is common in projectile motion under constant gravity.

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