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Chapter 3: Problem 70

Bang! \(\quad\) A student sits atop a platform a distance \(h\) above the ground. He throws a large firecracker horizontally with a speed \(v\) . However, a wind blowing parallel to the ground gives the fire- cracker a constant horizontal acceleration with magnitude \(a\) . This results in the firecracker reaching the ground directly under the student. Determine the height \(h\) in terms of \(v, a,\) and \(g\) . You can ignore the effect of air resistance on the vertical motion.

Short Answer

Expert verified
The height is \( h = \frac{2gv^2}{a^2} \).

Step by step solution

01

Analyze the Motion

The student throws the firecracker horizontally with initial horizontal velocity \(v\). A constant horizontal acceleration \(a\) is applied by the wind, affecting its horizontal motion. Vertically, the firecracker is in free fall, only affected by gravitational acceleration \(g\). The firecracker lands directly below the student's position, meaning there is no horizontal displacement. We need to determine the height \(h\).
02

Apply Kinematics to Horizontal Motion

Since the firecracker lands directly beneath the student, the net horizontal displacement is zero. The horizontal displacement \(x\) is given as \(x = vt + \frac{1}{2}at^2 = 0\). Therefore, we have:\[ vt + \frac{1}{2}at^2 = 0 \]and solving for \(t\), we find:\[ t = -\frac{2v}{a} \] (taking the non-trivial solution where \(v eq 0\) and \(a eq 0\)).
03

Apply Kinematics to Vertical Motion

Using the vertical kinematic equation for free fall, where the initial vertical velocity is zero, we have:\[ h = \frac{1}{2}gt^2 \]Substitute \(t = \frac{2v}{a}\) from Step 2 into this equation:\[ h = \frac{1}{2}g\left(\frac{2v}{a}\right)^2 \] which simplifies to:\[ h = \frac{2gv^2}{a^2} \].

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Key Concepts

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

Horizontal Displacement
In projectile motion, the concept of horizontal displacement is crucial. It refers to the distance traveled by a projectile along the horizontal axis during its flight. In this example, the firecracker experiences two opposing forces in the horizontal direction:
  • An initial velocity, denoted by \(v\).
  • A constant horizontal acceleration caused by the wind, denoted by \(a\).
The problem tells us that the firecracker lands directly beneath the student's position on the platform. This means that the total horizontal displacement, denoted as \(x\), is zero. Essentially, it moves forwards due to the initial throw and backwards because of the wind, effectively compensating each other.
The equation that describes this behavior is given as:
\[ x = vt + \frac{1}{2}at^2 = 0 \]
By solving this equation for time \(t\), we can understand how the forces cancel out, ultimately allowing us to find the time during which the firecracker is in flight.
Kinematic Equations
Kinematic equations provide us with a set of tools to describe the motion of objects under constant acceleration. They are applicable to both linear and projectile motion problems. In this exercise, we apply these equations to both horizontal and vertical motions separately.

For horizontal motion, we use:
\[ x = vt + \frac{1}{2}at^2 \]
This equation tells us how an object's position changes over time with initial velocity and constant acceleration. Since the horizontal displacement is zero, the equation simplifies to solve for time \(t\).

For vertical motion, we apply another kinematic equation:
\[ h = \frac{1}{2}gt^2 \]
Here, \(h\) represents the height from which the firecracker is dropped, and it is only affected by gravitational acceleration \(g\). Given the initial vertical velocity is zero, the height can be calculated by inserting the time \(t\) derived from the horizontal motion equation. This step allows us to connect the vertical fall to the time influenced by horizontal forces.
Gravitational Acceleration
Gravitational acceleration, denoted by \(g\), is a fundamental force that influences all objects near the Earth's surface, causing them to accelerate downwards at approximately 9.8 m/s虏. In this problem, only the vertical motion of the firecracker is affected by gravity.
The firecracker, once thrown horizontally, is subject to free fall. This results in a continuous increase in its vertical velocity as it descends towards the ground. Since the horizontal throw does not contribute to vertical motion, the initial vertical speed is zero.

Gravitational acceleration is reflected in the vertical kinematic equation:
\[ h = \frac{1}{2}gt^2 \]
This equation helps us determine the height \(h\) from which the object was dropped, based solely on the influence of gravity. By knowing how long the object was in the air (found from horizontal calculations), we can compute \(h\) by plugging in the known value of \(g\) and the calculated time \(t\). Understanding the interplay between time, gravity, and height is vital for solving vertical motion problems and highlights the intricacies of projectile motion.

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Most popular questions from this chapter

Leaping the River I. A car 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?

A shot putter releases the shot some distance above the level ground with a velocity of \(12.0 \mathrm{m} / \mathrm{s}, 51.0^{\circ}\) above the horizontal. The shot hits the ground 2.08 \(\mathrm{s}\) later. You can ignore air resistance. (a) What are the components of the shot's acceleration while in flight? (b) What are the components of the shot's velocity at the beginning and at the end of its trajectory? (c) How far did she throw the shot horizontally? (d) Why does the expression for \(R\) in Example 3.8 not give the correct answer for part (c)? (e) How high was the shot above the ground when she released it? Draw \(x-t\) , \(y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

The coordinstes of a bird flying in the \(x y\) -plane are given by \(x(t)=\alpha t\) and \(y(t)=3.0 \mathrm{m}-\beta t^{2},\) where \(\alpha=2.4 \mathrm{m} / \mathrm{s}\) and \(\beta=1.2 \mathrm{m} / \mathrm{s}^{2} \cdot\) (a) Sketch the path of the bird between \(t=0\) and \(t=2.0 \mathrm{s}\) . (b) Calculate the velocity and acceleration vectors of the bird as functions of time. (c) Calculate the magnitude and direction of the bird's velocity and acceleration at \(t=20 \mathrm{s}\) . (d) Sketch the velocity and acceleration vectors at \(t=2.0 \mathrm{s}\) . At this instant, is the bird speeding up, is it slowing down, or is its speed instantaneously not changing? Is the bird turning? If so, in what direction?

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 90.0 \(\mathrm{km} / \mathrm{h}\) , to his enemy's car, which is going 110 \(\mathrm{km} / \mathrm{h}\) . The enemy's car is 15.8 \(\mathrm{m}\) in front of the hero's when he lets go of the grenade. If the hero throws the grenade so its initial velocity relative to him is at an angle of \(45^{\circ}\) above the horizontal, what should the magnitude of the initial velocity be? The cars are both traveling in the same direction on a level road. You can ignore air resistance. Find the magnitude of the velocity both relative to the hero and relative to the earth.

Leaping the River I. A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Fig. 3.51\() .\) The takeoff ramp was inclined at \(53.0^{\circ}\) , the river was 40.0 \(\mathrm{m}\) wide, and the far bank was 15.0 \(\mathrm{m}\) lower than the top of the ramp. The river itself was 100 \(\mathrm{m}\) below the ramp. You can ignore air resistance. (a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? (b) If his speed was only half the value found in (a), where did he land?
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