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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 firecracker a constant horizontal acceleration with magnitude \(a\). As a result, the firecracker reaches the ground directly below the student. Determine the height \(h\) in terms of \(v\), \(a\), and \(g\). Ignore the effect of air resistance on the vertical motion.

Short Answer

Expert verified
The height \(h\) is given by \(h = \frac{2g v^2}{a^2}\).

Step by step solution

01

Analyze Horizontal Motion

In the horizontal direction, we'll use the equations of motion to analyze the effects of initial speed and acceleration due to wind. The initial horizontal velocity is \(v\) and the constant horizontal acceleration is \(a\). The time \(t\) taken for the firecracker to reach the ground can be found using: \[ x = vt + \frac{1}{2} a t^2 \]where \(x = 0\) because it directly lands below the starting point.
02

Solve for Time of Flight

Since \(x = 0\), the displacement in horizontal direction is zero. We can set up the equation:\[ 0 = v t + \frac{1}{2} a t^2 \]Simplifying this expression gives:\[ t (v + \frac{1}{2} a t) = 0 \]Since \(t eq 0\), we solve for \(t\): \[ v + \frac{1}{2} a t = 0 \]\[ t = -\frac{2v}{a} \]
03

Analyze Vertical Motion

Vertical motion is influenced only by gravity. We need to find the time it takes for the firecracker to hit the ground. Given that the initial vertical velocity is 0 and the gravitational acceleration is \(g\), the height \(h\) can be calculated using:\[ h = \frac{1}{2} g t^2 \]
04

Substitute Time into Vertical Motion Equation

Substitute the expression for time \(t = -\frac{2v}{a}\) from Step 2 into the vertical motion equation:\[ h = \frac{1}{2} g \left( -\frac{2v}{a} \right)^2 \]Simplify the expression:\[ h = \frac{1}{2} g \left( \frac{4v^2}{a^2} \right) \]\[ h = \frac{2g v^2}{a^2} \]

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Key Concepts

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

Horizontal Motion
When analyzing projectile motion, it's essential to understand horizontal motion. In this scenario, horizontal movement is influenced both by an initial velocity and a constant acceleration due to wind. The path of the firecracker is determined by its initial speed, denoted as \( v \), and the acceleration \( a \), which is parallel to the ground.

The horizontal displacement equation is:
  • \( x = vt + \frac{1}{2} a t^2 \)
For this problem, the firecracker lands directly beneath its starting point, making \( x = 0 \). This implies that the sum of the initial velocity multiplied by time and the product of acceleration and time squared (halved) results in no net displacement horizontally. This is crucial for finding the time of flight.
Vertical Motion
Vertical motion in projectile problems is primarily affected by gravitational forces. Here, the firecracker starts with an initial vertical velocity of zero, and gravity is the only force acting vertically on it.

Gravity accelerates the firecracker downwards at a rate of \( g \), the acceleration due to gravity. The core equation for vertical motion, given the starting parameters, is:
  • \( h = \frac{1}{2} g t^2 \)
This reveals that the height \( h \) from which the projectile was thrown is related to the time \( t \) it spends in the air and the gravitational pull.
Equations of Motion
Equations of motion are crucial for calculating and understanding the path of projectiles. In this exercise, we apply these equations both horizontally and vertically to discover various elements of motion.

For horizontal motion, where acceleration due to the wind is present, the equation:
  • \( x = vt + \frac{1}{2} a t^2 \)
is used to determine the duration of flight.

For vertical motion, the equation:
  • \( h = \frac{1}{2} g t^2 \)
links the time of flight to the height from which the projectile begins. Together, these equations help understand the firecracker's full trajectory.
Gravitational Acceleration
Gravitational acceleration, usually denoted by \( g \), is a constant force pulling objects toward the earth's center. This acceleration is approximately \( 9.8 \, \text{m/s}^2 \) on the surface of the Earth and plays a vital role in any type of projectile motion analysis.

In this scenario, since the firecracker is thrown horizontally, gravitational acceleration is the only force acting vertically. It dictates how long the firecracker is in the air by influencing the vertical motion equation:
  • \( h = \frac{1}{2} g t^2 \)
Through knowing \( g \), along with time determined from horizontal motion calculations, we can derive how far the firecracker falls, symbolized by the height \( h \). This intertwines gravitational forces with projectile motion to help predict outcomes accurately.

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Most popular questions from this chapter

A 76.0-kg rock is rolling horizontally at the top of a vertical cliff that is 20 m above the surface of a lake \(\textbf{(Fig. P3.65).}\) The top of the vertical face of a dam is located 100 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 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?

A railroad flatcar is traveling to the right at a speed of 13.0 m/s relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar (\(\textbf{Fig. E3.30}\)). What is the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter's velocity relative to the observer on the ground is (a) 18.0 m/s to the right? (b) 3.0 m/s to the left? (c) zero?

A sly 1.5-kg monkey and a jungle veterinarian with a blow-gun loaded with a tranquilizer dart are 25 m above the ground in trees 70 m apart. Just as the veterinarian shoots horizontally at the monkey, the monkey drops from the tree in a vain attempt to escape being hit. What must the minimum muzzle velocity of the dart be for the dart to hit the monkey before the monkey reaches the ground?

A man stands on the roof of a 15.0-m-tall building and throws a rock with a speed of 30.0 m/s at an angle of 33.0\(^\circ\) above the horizontal. Ignore air resistance. Calculate (a) the maximum height above the roof that the rock reaches; (b) the speed of the rock just before it strikes the ground; and (c) the horizontal range from the base of the building to the point where the rock strikes the ground. (d) Draw \(x-t, y-t, v_x-t\), and \(v_y-t\) graphs for the motion.

A toy rocket is launched with an initial velocity of 12.0 m/s in the horizontal direction from the roof of a 30.0-m-tall building. The rocket's engine produces a horizontal acceleration of \((1.60 m/s^3)t\), 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?

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