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Chapter 3: Problem 53

In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. A pilot is practicing by dropping a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a borizontal path 90.0 \(\mathrm{m}\) above the ground and with 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 274.56 meters from the target.

Step by step solution

01

Calculate Time of Flight

To determine when the canister hits the ground, compute the time it takes to fall 90.0 meters. Use the formula for free-fall motion: \[ d = \frac{1}{2}gt^2 \]where \(d = 90.0\, \mathrm{m}\) is the height, \(g = 9.8\, \mathrm{m/s^2}\) is the acceleration due to gravity, and \(t\) is the time in seconds.Rearranging the formula for \(t\), we get:\[ t = \sqrt{\frac{2d}{g}} \]Plugging in the values:\[ t = \sqrt{\frac{2 \times 90.0}{9.8}} \approx 4.29\, \mathrm{s} \]
02

Calculate Horizontal Distance

To find the horizontal distance where the canister should be released, use the formula:\[ x = vt \]where \(v = 64.0\, \mathrm{m/s}\) is the horizontal speed and \(t = 4.29\, \mathrm{s}\) is the time calculated in Step 1.Substitute the given values:\[ x = 64.0 \times 4.29 \approx 274.56\, \mathrm{m} \]
03

Conclusion

The pilot should release the canister approximately 274.56 meters from the target to ensure it lands on the target.

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Key Concepts

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

Free Fall
Free fall refers to the motion of an object under the influence of gravity alone, without any resistance like air drag. It is a simple yet vital concept in understanding projectile motion.
When an object is in free fall, the only force acting on it is gravity. This means it accelerates downward at a constant rate known as the acceleration due to gravity, which is approximately 9.8 meters per second squared (\( g = 9.8 \, \mathrm{m/s^2} \)). In the exercise provided, the canister dropped by the airplane exhibits free fall once released.
Key points to remember about free fall:
  • The initial vertical velocity of the object is usually zero as it's dropped from rest or released into free fall.
  • The only acceleration is due to gravity, so no other forces are involved.
  • This concept simplifies the calculations related to the time it takes an object to fall a certain distance.
Horizontal Distance
Horizontal distance in projectile motion refers to the distance an object travels along the horizontal axis during its motion. It is crucial in situations like the exercise where a canister must be dropped to hit a target.
To calculate the horizontal distance, use the formula:
\[ x = vt \]
where \(v\) is the constant horizontal speed of the plane, and \(t\) is the time of flight calculated from the free fall part of the motion.
Important aspects of horizontal distance:
  • The speed in the horizontal direction remains constant if we ignore air resistance, as stated in the exercise.
  • The time of flight, determined by the object's vertical motion, is used to compute how far the object travels horizontally from the release point.
  • In the given example, after calculating the time \(t\), the horizontal distance \(x\) is found to ensure the canister reaches the target.
Acceleration due to Gravity
The acceleration due to gravity is a fundamental concept in physics that affects all free-falling objects on Earth. Denoted by \( g \), its standard value is approximately 9.8 meters per second squared (\( \mathrm{m/s^2} \)). This value indicates the rate at which an object's velocity changes when it is in free fall.
Understanding acceleration due to gravity is essential when analyzing projectile motion.
Features of the acceleration due to gravity include:
  • Constant value: On Earth, \( g \) remains fairly constant, simplifying calculations for free-falling objects.
  • Direction: Always acts vertically downward, towards the center of the Earth.
  • Impact on motion: Determines how quickly an object speeds up while descending and is a critical factor in determining time of flight.
In the exercise, this acceleration was used to calculate the time it takes for the canister to reach the ground, impacting where it should be released to strike the target.

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Most popular questions from this chapter

Dynamite! A demolition crew uses dynamite to blow an old building apart. Debris from the explosion fies off in all directions and is later found at distances as far as 50 \(\mathrm{m}\) from the explosion. Estimate the maximum speed at which debris was blown outward by the explosion. Describe any assumptions that you make.

A military helicopter on a training mission is flying horizontally at a speed of 60.0 \(\mathrm{m} / \mathrm{s}\) and accidentally drops a bomb (fortunately not armed) at an elevation of 300 \(\mathrm{m} .\) You can ignore air resistance. (a) How much time is required for the bomb to reach the earth? (b) How far does it travel horizontally while falling? (c) Find the horizontal and vertical components of its velocity just before it strikes the earth. (d) Draw \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the bomb's motion. (e) If the velocity of the helicopter remains con0stant, where is the helicopter when the bomb hits the ground?

Cycloid. A particle moves in the \(x y\) -plane. Its coordinates are given as functions of time by $$ x(t)=R(\omega t-\sin \omega t) \quad y(t)=R(1-\cos \omega t) $$ where \(R\) and \(\omega\) are constants. (a) Sketch the trajectory of the particle. (This is the trajectory of a point on the rim of a wheel that is rolling at a constant speed on a horizontal surface. The curve traced out by such a point as it moves through space is called a cycloid.) (b) Determine the velocity components and the acceleration components of the particle at any time \(t\) . (c) At which times is the particle momentarily at rest? What are the coordinates of the particle at these times? What are the magnitude and direction of the acceleration at these times? (d) Does the magnitude of the acceleration depend on time? Compare to uniform circular motion.

An airplane is flying with a velocity of \(90.0 \mathrm{m} / \mathrm{s}\) at an angle of \(23.0^{\circ}\) above the horizontal. When the plane is \(114 \mathrm{m}\) directly above a dog that is standing on level ground, a suitcase drops out of the Iuggage compartment. How far from the dog will the suitcase land? You can ignore air resistance.

A dog running in an open field has components of velocity \(\boldsymbol{v}_{x}=2.6 \mathrm{m} / \mathrm{s}\) 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?
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