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Chapter 3: Problem 54

An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at 150 m above the level ground when the plane is flying at 75 m/s in a direction 55\(^{\circ}\) above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales land at the point where the cattle are stranded?

Short Answer

Expert verified
The pilot should release the hay 649 meters in front of the cattle.

Step by step solution

01

Analyze the Problem

We need to calculate the horizontal distance from the point of release to where the bales land. Use projectile motion because the bales are affected by gravity after being dropped.
02

Break Down Initial Velocity

Find the horizontal and vertical components of the initial velocity using trigonometry. Given the initial speed of 75 m/s, the horizontal component is \(v_{x} = 75 \cos(55^{\circ})\) and the vertical component is \(v_{y} = 75 \sin(55^{\circ})\).
03

Calculate Time of Flight

Use the vertical motion to find the time it takes for the bales to hit the ground. The initial vertical position is 150 m. Use the equation: \[ y = v_{y}t + \frac{1}{2}gt^2 \]where \(g = -9.81 \text{ m/s}^2\), solve for \(t\) when \(y = 0\).
04

Solve for Time of Flight

Rearrange and solve the quadratic equation for time:\[ -4.905t^2 + 75 \sin(55^{\circ})t - 150 = 0 \].Use the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find \(t\).
05

Find the Horizontal Distance

With the time of flight \(t\), calculate the horizontal distance the bales travel using the horizontal velocity. Use:\( x = v_{x} \times t \), where \(v_{x} = 75 \cos(55^{\circ})\).
06

Final Calculation and Answer

Perform all calculations: 1. Calculate \(v_{x} \approx 43.0 \text{ m/s}\) and \(v_{y} \approx 61.4 \text{ m/s}\).2. Solve the quadratic equation to find \(t \approx 15.1 \text{ seconds}\).3. Calculate \(x = 43.0 \times 15.1 \approx 649 \text{ meters}\).

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Key Concepts

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

Initial Velocity Components
When dealing with projectile motion, it's essential to break down the initial velocity into horizontal and vertical components. This allows us to analyze the motion separately in each direction.
For a projectile launched at an angle, the overall initial velocity (v) can be divided into two parts using simple trigonometry: the horizontal velocity (v_x) and the vertical velocity (v_y). This involves:
  • Using cosine for horizontal motion: \( v_x = v \cos(\theta) \)
  • Using sine for vertical motion: \( v_y = v \sin(\theta) \)
For our problem, where the airplane releases the bales with an initial speed of 75 m/s at 55掳,
  • The horizontal component: \( v_x = 75 \cos(55^\circ) \approx 43.0 \) m/s
  • The vertical component: \( v_y = 75 \sin(55^\circ) \approx 61.4 \) m/s
Understanding these components is critical because they tell us how fast and in what direction the bales will initially start moving in both dimensions.
Time of Flight
The time of flight is the total time the bales are in the air, from the moment they leave the plane to when they hit the ground.
To calculate this, we focus on the vertical motion, since gravity only affects the projectile vertically.An important formula used here is the equation for vertical displacement:\[ y = v_y t + \frac{1}{2} g t^2\]Where:
  • \( y \) is the vertical displacement, initially 150 m in our case.
  • \( v_y \) is the initial vertical velocity.
  • \( g \) is the acceleration due to gravity, \( g = -9.81 \text{ m/s}^2 \).
For the bales to hit the ground from a height of 150 m, we solve when \( y = 0 \).This results in a quadratic equation:\[ -4.905t^2 + 61.4t - 150 = 0\]Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we solve for \( t \), finding that the time of flight is approximately \( 15.1 \) seconds.
Horizontal Distance Calculation
Once we know how long the bales are in the air, calculating the horizontal distance traveled is straightforward. The horizontal motion is constant because there's no horizontal acceleration acting on the bales.
The formula to find the horizontal distance (\( x \)) is given by:\[ x = v_x \times t\]Where:
  • \( v_x \) is the horizontal component of the initial velocity, which we previously found to be \( 43.0 \text{ m/s} \)
  • \( t \) is the time of flight, \( 15.1 \) seconds.
By substituting these values into the equation:\[ x = 43.0 \times 15.1 \approx 649 \text{ meters}\]Thus, the pilot should release the bales approximately 649 meters before reaching the cattle, ensuring they land right where needed. Understanding this concept helps solve similar problems in projectile motion, ensuring a clear grasp of how motion progresses in two dimensions.

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Most popular questions from this chapter

On level ground a shell is fired with an initial velocity of 40.0 m/s at 60.0\(^\circ\) above the horizontal and feels no appreciable air resistance. (a) Find the horizontal and vertical components of the shell's initial velocity. (b) How long does it take the shell to reach its highest point? (c) Find its maximum height above the ground. (d) How far from its firing point does the shell land? (e) At its highest point, find the horizontal and vertical components of its acceleration and velocity.

CALC A web page designer creates an animation in which a dot on a computer screen has position $$ \vec{r} =[34.0 cm +(2.5 cm/s^2)t^2] \hat{i} +(5.0 cm/s)t \hat{\jmath}.$$ (a) Find the magnitude and direction of the dot's average velocity between \(t\) = 0 and \(t\) = 2.0 s.(b) Find the magnitude and direction of the instantaneous velocity at \(t\) = 0, \(t\) = 1.0 s, and \(t\) = 2.0 s. (c) Sketch the dot's trajectory from \(t\) = 0 to \(t\) = 2.0 s, and show the velocities calculated in part (b).

Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 km/h. If one goose is flying at 100 km/h relative to the air but a 40-km/h wind is blowing from west to east, (a) at what angle relative to the north-south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of 500 km from north to south? (\(Note\): Even on cloudy nights, many birds can navigate by using the earth's magnetic field to fix the north-south direction.)

During a storm, a car traveling on a level horizontal road comes upon a bridge that has washed out. The driver must get to the other side, so he decides to try leaping the river with his car. The side of the road the car is on is 21.3 m above the river, while the opposite side is only 1.8 m above the river. The river itself is a raging torrent 48.0 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 major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an angle of 36.9\(^\circ\) above the horizontal. Ignore air resistance. (a) At what \(two\) times is the baseball at a height of 10.0 m above the point at which it left the bat? (b) Calculate the horizontal and vertical components of the baseball's velocity at each of the two times calculated in part (a). (c) What are the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat?
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