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Chapter 3: Problem 53

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 \(30.0-\mathrm{m}\) -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. Ignore air resistance. What horizontal distance does the rocket travel before reaching the ground?

Short Answer

Expert verified
To find the horizontal distance travelled by the rocket before reaching the ground, bind the separate horizontal and vertical motions, solve these separately using the equations of motion, substitute the calculated time of flight into the horizontal distance equation and calculate the result.

Step by step solution

01

Identify known and unknown variables

From the problem, it is known that the toy rocket is launched with an initial horizontal velocity \(v_{ix} = 12.0 m/s\), vertical displacement \(d_y = -30.0 m\) (negative because it's downward), horizontal acceleration \(a_x = 1.60 m/s^2 t\), and vertical acceleration \(a_y = g\). The problem asks for finding the horizontal distance traveled by the rocket before reaching the ground, which implies \(d_x\).
02

Find the time required for the rocket to hit the ground

First, look into the vertical motion. When the object is in a free fall, it can be represented by the equation \(d_y = v_{iy}t - \frac{1}{2}gt^2\). Since the rocket was only launched horizontally, \(v_{iy} = 0\). Thus, the equation can be simplified to \(d_y = - \frac{1}{2}gt^2\). Solving for \(t\), we can get guess the rocket's total time of flight, \(t = \sqrt{-2d_y/g}\).
03

Calculate the horizontal distance travelled

Now, look into the horizontal motion. Given the horizontal acceleration of the rocket is variable, the distance can be calculated as \(d_x = v_{ix} t + \frac{1}{2} a_x t^2\). Substituting the value of \(t\) from step 2 and the given value of \(a_x\), calculate the \(d_x\).
04

Verify the units

As a final step, verify the unit of the obtained distance. It should be in metres (m).

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Key Concepts

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

Kinematics in Two Dimensions
Understanding kinematics in two dimensions is essential for analysing movements like a toy rocket being propelled in the air. In two-dimensional motion, objects have two independent movements: along the horizontal axis (x-axis) and vertical axis (y-axis). These movements are described by separate sets of equations that consider velocity, acceleration, and displacement.

In the case of the toy rocket problem, we deal with the rocket moving horizontally off the building with an initial velocity and an additional horizontal acceleration due to the engine thrust. Vertically, it begins without initial velocity but experiences the constant acceleration due to gravity. To solve such problems, we separately analyze the motion along each axis and then combine the results to determine the resultant path or final position.
Free Fall Acceleration
Free fall acceleration, typically represented by the symbol \(g\), is the constant acceleration experienced by an object when it is falling solely under the influence of gravity. The value of \(g\) approximates to \(9.81 \text{m/s}^2\) on Earth's surface. Free fall acceleration acts in the downward vertical direction and is a critical factor in determining the time and impact velocity of falling objects.

For the toy rocket scenario, once it leaves the initial thrust of its propulsion, it will accelerate downwards at this rate. It's important to note that the toy rocket's horizontal motion is unaffected by gravity and we consider it only for its vertical motion.
Equations of Motion
The equations of motion let us predict future motion based on current and past motion conditions. For constant acceleration, they can be written as:
  • \( v = v_0 + a t \)
  • \( d = v_0 t + \frac{1}{2} a t^2 \)
  • \( v^2 = v_0^2 + 2 a d \)
where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(a\) is the acceleration, \(t\) is the time, and \(d\) is the displacement.

For instance, while the horizontal motion of our toy rocket includes an accelerating force from the engine, leading to a variable acceleration, the vertical motion is subject to a consistent acceleration due to gravity. Therefore, we use the appropriate motion equations for each dimension when solving for the time of flight and the horizontal distance traveled.
Time of Flight
The term 'time of flight' refers to the duration an object stays in the air. Determining the time of flight for projectile motion is a two-step process: first, we find the time it takes for the object to reach the peak of its trajectory, and second, we find the time it takes to fall back down to its original elevation or to the ground if it starts from a height. However, if an object launches horizontally from a height, like our toy rocket, we only need to consider the descent.

To find the time of flight for our horizontally launched rocket, we can ignore the horizontal motion and just consider the vertical fall. Using the equation \( t = \text{sqrt}{(-2d_y/g)} \), with \( d_y = -30.0 \text{m} \) due to gravity and the height the rocket falls from, allows us to compute how long the rocket will be in the air before it impacts the ground.

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Most popular questions from this chapter

A \(76.0 \mathrm{~kg}\) rock is rolling horizontally at the top of a vertical cliff that is \(20 \mathrm{~m}\) above the surface of a lake (Fig. \(\mathrm{P} 3.65\) ). The top of the vertical face of a dam is located \(100 \mathrm{~m}\) from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is \(25 \mathrm{~m}\) below the top of the dam. (a) What must be the minimum speed of the rock just as it leaves the cliff so that it will reach the plain without striking the dam? (b) How far from the foot of the dam does the rock hit the plain?

When a train's velocity is \(12.0 \mathrm{~m} / \mathrm{s}\) eastward, raindrops that are falling vertically with respect to the earth make traces that are inclined \(30.0^{\circ}\) to the vertical on the windows of the train. (a) What is the horizontal component of a drop's velocity with respect to the earth? With respect to the train? (b) What is the magnitude of the velocity of the raindrop with respect to the earth? With respect to the train?

A small object is projected from level ground with an initial velocity of magnitude \(16.0 \mathrm{~m} / \mathrm{s}\) and directed at an angle of \(60.0^{\circ}\) above the horizontal. (a) What is the horizontal displacement of the object when it is at its maximum height? How does your result compare to the horizontal range \(R\) of the object? (b) What is the vertical displacement of the object when its horizontal displacement is \(80.0 \%\) of its horizontal range \(R ?\) How does your result compare to the maximum height \(h_{\max }\) reached by the object? (c) For when the object has horizontal displacement \(x-x_{0}=\alpha R,\) where \(\alpha\) is a positive constant, derive an expression (in terms of \(\alpha\) ) for \(\left(y-y_{0}\right) / h_{\max }\). Your result should not depend on the initial velocity or the angle of projection. Show that your expression gives the correct result when \(\alpha=0.80,\) as is the case in part (b). Also show that your expression gives the correct result for \(\alpha=0, \alpha=0.50\), and \(\alpha=1.0\)

In a World Cup soccer match, Juan is running due north toward the goal with a speed of \(8.00 \mathrm{~m} / \mathrm{s}\) relative to the ground. A teammate passes the ball to him. The ball has a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) and is moving in a direction \(37.0^{\circ}\) east of north, relative to the ground. What are the magnitude and direction of the ball's velocity relative to Juan?

A firefighting crew uses a water cannon that shoots water at \(25.0 \mathrm{~m} / \mathrm{s}\) at a fixed angle of \(53.0^{\circ}\) above the horizontal. The firefighters want to direct the water at a blaze that is \(10.0 \mathrm{~m}\) above ground level. How far from the building should they position their cannon? There are \(t w o\) possibilities; can you get them both? (Hint: Start with a sketch showing the trajectory of the water.
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