/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 In fighting forest fires, airpla... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 49

In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. For practice, a pilot drops a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path \(90.0 \mathrm{~m}\) above the ground and has a speed of \(64.0 \mathrm{~m} / \mathrm{s}(143 \mathrm{mi} / \mathrm{h}),\) at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.

Short Answer

Expert verified
The pilot should release the canister approximately 273.92 m from the target to hit it accurately.

Step by step solution

01

Calculate the Fall Time

The first step is to calculate how long the canister takes to fall. This can be done using the formula for the time it takes an object to fall from a certain height under gravity: \( t = \sqrt{\frac{2h}{g}} \), where \( h = 90.0 \, \mathrm{m} \) is the altitude of the plane and \( g = 9.81 \, \mathrm{m/s^2} \) is the acceleration due to gravity.
02

Compute the Fall Time

Substitute \( h = 90.0 \, \mathrm{m} \) and \( g = 9.81 \, \mathrm{m/s^2} \) into the formula to find the time : \( t = \sqrt{\frac{2 * 90.0 \, \mathrm{m}}{9.81 \, \mathrm{m/s^2}}} \). The result is approximately \( t = 4.28 \, \mathrm{s} \).
03

Calculate the Horizontal Distance

Now, use the speed of the plane and the time it takes for the canister to fall to calculate the horizontal distance from the target where the canister should be dropped. Since the horizontal distance is the product of speed and time, use the equation \( d = vt \), where \( v = 64.0 \, \mathrm{m/s} \) is the speed of the plane and \( t = 4.28 \, \mathrm{s} \) is the time it takes for the canister to fall. So, \( d = (64.0 \, \mathrm{m/s}) * (4.28 \, \mathrm{s}) \).

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Key Concepts

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

gravity
When discussing projectile motion, gravity plays a crucial role. It is the force that pulls objects toward the Earth's center, causing them to fall when dropped or thrown. In the given problem, we are dealing with a canister dropped from an airplane. Since gravity is pulling it towards the ground, it becomes a vital factor in calculating the time it takes for the canister to hit the ground from a height of 90 meters.
To understand the effects of gravity, we use the formula for an object's fall time: \( t = \sqrt{\frac{2h}{g}} \), where \( h \) is the height (90 meters in this case), and \( g \) is the acceleration due to gravity, which is approximately 9.81 m/s虏 on Earth. This formula helps to determine how long the canister will take to fall, which is essential for deciding when to release it from the airplane. Without considering gravity, the canister's path to the ground would be miscalculated.
horizontal distance
In projectile motion, horizontal distance refers to how far an object travels along the horizontal plane before hitting the ground. For our canister dropped from an airplane, calculating this is essential to ensure it hits the target on the ground. The key to determining horizontal distance is understanding that, in the absence of air resistance, horizontal velocity remains constant.
The formula we use here is \( d = vt \), where \( d \) is the horizontal distance, \( v \) is the horizontal velocity (64 m/s for the airplane), and \( t \) is the time calculated previously (4.28 seconds). By substituting these values, you find the horizontal distance the canister travels before touching the ground. This precise calculation allows the pilot to time the release optimally.
vertical motion
Vertical motion involves the movement of an object as it falls under the influence of gravity, excluding any initial vertical velocity, since our problem assumes a horizontal drop. This means that the only force acting vertically on the canister is gravity, causing it to accelerate downward.
In physics, this vertical fall is crucial for calculating when and how objects will hit the ground. For the canister, knowing the time it takes to fall (calculated using gravitational formulas) informs both when the object should be released and how gravity impacts its descent. Studying vertical motion helps to understand the downward trajectory and ensures our canister lands precisely at the target location. Get familiar with these vertical concepts to master relating them to real-world physics problems like this.

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Most popular questions from this chapter

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?

A projectile thrown from a point \(P\) moves in such a way that its distance from \(P\) is always increasing. Find the maximum angle above the horizontal with which the projectile could have been thrown. Ignore air resistance.

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 \(t_{1}=10.0 \mathrm{~s}\) to \(t_{2}=20.0 \mathrm{~s},\) 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?

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?

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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