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Chapter 3: Problem 57

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

Short Answer

Expert verified
The horizontal distance before hitting the ground is 59.23 meters.

Step by step solution

01

Analyze the problem

The toy rocket is moving horizontally with an initial speed of 12.0 m/s from a height of 30.0 m. The horizontal acceleration of the rocket changes with time, given by \( a(t) = 1.60t \). Vertically, the acceleration is due to gravity, \( g = 9.81 \text{ m/s}^2 \). We need to find how far the rocket travels horizontally until it hits the ground.
02

Calculate time of flight

Using the vertical motion, where the initial vertical velocity is 0 m/s, we solve for time \( t \) when the rocket hits the ground using the equation: \( y = \( v_{0y} \) t + \frac{1}{2} a_y t^2 \) with \( y = 0 - 30 \), \( v_{0y} = 0 \), and \( a_y = -9.81 \text{ m/s}^2 \). This gives us: \[ 0 - 30 = 0 + \frac{1}{2}(-9.81)t^2 \]\[ -30 = -4.905t^2 \]Solving for \( t \), we find \( t = \sqrt{ \frac{30}{4.905} } \approx 2.47 \text{ seconds} \).
03

Calculate horizontal distance

The horizontal distance \( x \) is found using the equation for horizontal motion: \[ x = v_{0x}t + \frac{1}{2} a(t)t^2 \]Substitute \( v_{0x} = 12.0 \text{ m/s} \), \( a(t) = 1.60t \), and \( t = 2.47 \text{ seconds} \):\[ x = 12.0(2.47) + \frac{1}{2}(1.60)(2.47)^2(2.47) \]Calculate:\[ x = 12.0(2.47) + 0.8(14.94)(2.47) \]\[ x = 29.64 + 29.59 \]\[ x = 59.23 \text{ meters} \].

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Key Concepts

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

Horizontal Velocity
When a toy rocket is launched, it can have a horizontal velocity, which affects how far it will travel before descending. Horizontal velocity is the speed at which the rocket moves sideways. For the rocket in this problem,the initial horizontal velocity is given as 12.0 m/s.

Unlike vertical motion, horizontal motion in this case doesn't have an external force affecting its initial velocity directly, except for a changing horizontal accelerationalready provided in the equation as a function of time.
  • Initial horizontal velocity: 12.0 m/s.
  • Effect of horizontal acceleration: Increases the horizontal velocity over time.
The horizontal velocity changes due to the acceleration provided by the engine, which is expressed as \( 1.60t \) m/s虏. This means as time increases, the horizontal component of the rocket's motion increases linearly with time, boosting the distance traveled.
Vertical Acceleration
While the rocket travels horizontally, it also experiences vertical forces. Since there is no initial vertical velocity,the only vertical motion the rocket experiences is due to gravity. Gravity accelerates the rocket downward at a constant rate.

For this exercise, the vertical acceleration is equivalent to the acceleration due to gravity, \( 9.81 \text{ m/s}^2 \). However, because the gravity force acts downwards, it's treated as a negative acceleration when calculating the time of flight.
  • Vertical acceleration due to gravity: \(-9.81 \text{ m/s}^2\).
  • No initial vertical velocity: Starts from rest vertically.
Understanding vertical acceleration is crucial to figure out how long the rocket takes to hit the ground,because it defines how fast the rocket speeds up as it falls.
Time of Flight
The time of flight is the duration the rocket remains airborne before hitting the ground. This duration determines both how far it travels horizontally and how fast it falls vertically.

To find the time of flight, we consider the vertical motion, with an initial vertical position of 0 m/s. We use the equation for vertical displacement: \[ y = v_{0y}t + \frac{1}{2} a_y t^2 \].

In this scenario:
  • Start height: 30 m (negative because it falls from the ground to the building top).
  • Initial vertical velocity: 0 m/s.
  • Vertical acceleration: \(-9.81 \text{ m/s}^2\).
By substituting these values into the equation, the time \( t \) is calculated as approximately 2.47 seconds. This is how long the rocket spends traveling from the top of the building to the ground.
Horizontal Distance
Once the time of flight is determined, the next step is to calculate the horizontal distance the rocket covers. The horizontal distance depends on both the initial horizontal velocity and the changing acceleration over time.

The formula used is \[ x = v_{0x}t + \frac{1}{2} a(t)t^2 \], where:
  • \( v_{0x} \) is the initial horizontal velocity (12.0 m/s).
  • \( t \) is the time of flight (2.47 seconds).
  • \( a(t) = 1.60t \) is the time-varying horizontal acceleration.
The horizontal distance thus incorporates the effects of initial speed and the extra speed gained due to the linear increase in horizontal acceleration. After calculating using these variables, the horizontal distance traveled by the rocket is 59.23 meters. This computation shows how integrating changes in velocity and acceleration over time gives us the full picture of the rocket's horizontal journey.

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Most popular questions from this chapter

A physics book slides off a horizontal tabletop with a speed of 1.10 \(\mathrm{m} / \mathrm{s} .\) It strikes the floor in 0.350 s. Ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor. (d) Draw \(x-t, y-t, v_{x^{-}},\) and \(v_{y}-t\) graphs for the motion.

A car traveling on a level horizontal road comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side of the road the car is on is 21.3 \(\mathrm{m}\) above the river, while the opposite side is a mere 1.8 \(\mathrm{m}\) above the river. The river itself is a raging torrent 61.0 \(\mathrm{m}\) wide. (a) How fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side? (b) What is the speed of the car just before it lands on the other side?

An airplane pilot wishes to fly due west. A wind of 80.0 \(\mathrm{km} / \mathrm{h}(\) about 50 \(\mathrm{mi} / \mathrm{h})\) is blowing toward the south. (a) If the airspeed of the plane (its speed in still air) is 320.0 \(\mathrm{km} / \mathrm{h}\) (about 200 \(\mathrm{mi} / \mathrm{h} )\) , in which direction should the pilot head? (b) What is the speed of the plane over the ground? Illustrate with a vector diagram.

Two crickets, Chirpy and Milada, jump from the top of a vertical cliff. Chirpy just drops and reaches the ground in 3.50 s, while Milada jumps horizontally with an initial speed of 95.0 \(\mathrm{cm} / \mathrm{s} .\) How far from the base of the cliff will Milada hit the ground?

A dog running in an open field has components of velocity \(v_{x}=2.6 \mathrm{m} / \mathrm{s}\) and \(v_{y}=-1.8 \mathrm{m} / \mathrm{s}\) at \(t_{1}=10.0 \mathrm{s}\) . For the time interval from \(t_{1}=10.0\) s to \(t_{2}=20.0\) s, the average acceleration of the dog has magnitude 0.45 \(\mathrm{m} / \mathrm{s}^{2}\) and direction \(31.0^{\circ}\) measured from the \(+x\) -axis toward the \(+y\) -axis. At \(t_{2}=20.0 \mathrm{s}\) (a) what are the \(x\) - and \(y\) -components of the dog's velocity? (b) What are the magnitude and direction of the dog's velocity? (c) Sketch the velocity vectors at \(t_{1}\) and \(t_{2} .\) How do these two vectors differ?
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