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\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?

Short Answer

Expert verified
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 by step solution

01

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.
02

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):\[ s = 40.0 \times 4.08 + \frac{1}{2} \times (-9.8) \times (4.08)^2 \]\[ s = 163.2 - 81.6 \]\[ s = 81.6 \text{ meters} \]
03

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):\[ d = 30.0 \times 8.16 \]\[ d = 244.8 \text{ meters} \]
04

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

\(\bullet\) Bird migration. Canadian geese migrate essentially along a north- south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 \(\mathrm{km} / \mathrm{h}\) . If one such bird is flying at 100 \(\mathrm{km} / \mathrm{h}\) relative to the air, but there is a 40 \(\mathrm{km} / \mathrm{h}\) wind blowing from west to east, (a) at what angle relative to the north-south direction should this bird head so that it will be traveling directly southward relative to the ground? (b) How long will it take the bird to cover a ground distance of 500 \(\mathrm{km}\) from north to south? (Note: Even on cloudy nights, many birds can navigate using the earth's magnetic field to fix the north-south direction.)

A meteor streaking through the night sky is located with radar. At point \(A\) its coordinates are \((5.00 \mathrm{km}, 1.20 \mathrm{km}),\) and 1.14 s later it has moved to point \(B\) with coordinates \((6.24 \mathrm{km},\)0.925 \(\mathrm{km} ) .\) Find (a) the \(x\) and \(y\) components of its average velocity between \(A\) and \(B\) and (b) the magnitude and direction of its average velocity between these two points.

\(\bullet\) A player kicks a football at an angle of \(40.0^{\circ}\) from the horizontal, with an initial speed of 12.0 \(\mathrm{m} / \mathrm{s} .\) A second player standing at a distance of 30.0 \(\mathrm{m}\) from the first (in the direction of the kick) starts running to meet the ball at the instant it is kicked. How fast must he run in order to catch the ball just before it hits the ground?

If the time of flight of the ball is \(t\) seconds, at what point in time will the ball have zero vertical velocity? A. \(t / 4 \mathrm{s}\) B. \(t / 4 \mathrm{s}\) C. \(t / 2 \mathrm{s}\) D. \(t / 2 \mathrm{s}\) E. There is no place on the path where the vertical velocity is zero.

A military helicopter on a training mission is flying horizontally at a speed of 60.0\(\mathrm{m} / \mathrm{s}\) when it accidentally drops a bomb(fortunately, not armed) at an elevation of 300 \(\mathrm{m.}\) You can ignore air resistance. (a) How much time is required for the bomb to reach the earth? (b) How far does it travel horizontally while falling? (c) Find the horizontal and vertical components of the bomb's velocity just before it strikes the earth. (d) Draw graphs of the horizontal distance vs. time and the vertical distance vs. time for the bomb's motion. (e) If the velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground?

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