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Chapter 2: Problem 27

A toy helicopter is flying in a straight line at a constant speed of \(4.5 \mathrm{m} / \mathrm{s}\). If a projectile is launched vertically with an initial speed of \(v_{0}=28 \mathrm{m} / \mathrm{s}\), what horizontal distance \(d\) should the helicopter be from the launch site \(S\) if the projectile is to be traveling downward when it strikes the helicopter? Assume that the projectile travels only in the vertical direction.

Short Answer

Expert verified
The helicopter should be approximately 12.86 meters from the launch site.

Step by step solution

01

Identify the Time of Flight for the Projectile

The first step is to calculate the time at which the projectile is traveling downward. The motion of the projectile is governed by the equation for velocity: \[ v = v_0 - g t \] where \( v_0 = 28 \, \text{m/s} \) is the initial velocity, \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( v \) is the velocity when the projectile is traveling downward. Since it must be traveling downward with a zero upward component (meaning it just starts falling down), we set \( v = 0 \):\[ 0 = 28 \, \text{m/s} - 9.8 \, \text{m/s}^2 \cdot t \]Solve for \( t \):\[ t = \frac{28}{9.8} \approx 2.857 \text{ s} \]
02

Calculate Horizontal Distance Traveled by the Helicopter

Knowing the projectile strikes the helicopter at approximately \( 2.857 \) seconds, we use the helicopter's speed to calculate the distance it has traveled. The helicopter flies horizontally at a constant speed of \( 4.5 \, \text{m/s} \). Therefore, the horizontal distance \( d \) traveled by the helicopter is given by:\[ d = \text{speed} \times \text{time} = 4.5 \, \text{m/s} \times 2.857 \, \text{s} \approx 12.86 \, \text{m} \]
03

Conclusion

The horizontal distance \( d \) that the helicopter should be from the launch site \( S \) for the projectile to strike it while traveling downward is approximately \( 12.86 \, \text{m} \).

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Key Concepts

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

Constant Speed
When we talk about constant speed, it means the object is moving at a steady rate without speeding up or slowing down. In our exercise, the toy helicopter flies horizontally at a constant speed of 4.5 meters per second. This constant speed is crucial because it allows us to predict the helicopter's position at any given time.
  • Constant speed = No change in speed over time
  • The helicopter covers the same amount of distance each second
Knowing the speed is vital as it helps to calculate how far the helicopter will travel in the time it takes for the projectile to meet it.
Vertical Velocity
Vertical velocity is the speed at which an object moves up or down. In the problem, the projectile is launched vertically with an initial speed of 28 meters per second. The key point here is that acceleration due to gravity affects this vertical motion, slowing it down until it stops rising and begins to fall.
  • Initial vertical velocity: 28 m/s
  • Gravity decreases this speed until the projectile starts falling down
At the instant when the vertical velocity becomes zero, the projectile is at its highest point. After this, it begins to travel downward with an increasing speed.
Horizontal Distance
Horizontal distance refers to how far horizontally an object travels. For the helicopter, this is calculated using its constant speed multiplied by the time duration.
  • Speed of helicopter: 4.5 m/s
  • Time taken for projectile to reach the apex and start down: approximately 2.857 seconds
The horizontal distance, therefore, becomes a straightforward multiplication problem: speed times time equals distance. In this case, it鈥檚 approximately 12.86 meters from point S to where it would meet the descending projectile.
Acceleration Due to Gravity
Acceleration due to gravity is a constant force pulling objects towards the Earth. Its value is approximately 9.8 meters per second squared. This force impacts the projectile by steadily decreasing its upward speed until it starts to fall back down.
  • Acceleration due to gravity: 9.8 m/s虏
  • Always acts downward
This unchanging force helps us determine how long the projectile takes to stop rising and start descending. By knowing this, we can precisely calculate the timing of the projectile鈥檚 flight and ensure it aligns with the helicopter鈥檚 path.

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Most popular questions from this chapter

Determine the vertical rise \(h\) of the load \(W\) during 10 seconds if the hoisting drum draws in cable at the constant rate of \(180 \mathrm{mm} / \mathrm{s}\)

An electric motor \(M\) is used to reel in cable and hoist a bicycle into the ceiling space of a garage. Pulleys are fastened to the bicycle frame with hooks at locations \(A\) and \(B\), and the motor can reel in cable at a steady rate of 12 in./sec. At this rate, how long will it take to hoist the bicycle 5 feet into the air? Assume that the bicycle remains level.

The stories of a tall building are uniformly 10 feet in height. A ball \(A\) is dropped from the rooftop position shown. Determine the times required for it to pass the 10 feet of the first, tenth, and onehundredth stories (counted from the top). Neglect aerodynamic drag.

A block of mass \(m\) rests on a rough horizontal surface and is attached to a spring of stiffness \(k .\) The coefficients of both static and kinetic friction are \(\mu\) The block is displaced a distance \(x_{0}\) to the right of the unstretched position of the spring and released from rest. If the value of \(x_{0}\) is large enough, the spring force will overcome the maximum available static friction force and the block will slide toward the unstretched position of the spring with an acceleration \(a=\mu g-\frac{k}{m} x,\) where \(x\) represents the amount of stretch (or compression) in the spring at any given location in the motion. Use the values \(m=5 \mathrm{kg}, k=150 \mathrm{N} / \mathrm{m}, \mu=0.40,\) and \(x_{0}=200 \mathrm{mm}\) and determine the final spring stretch (or compression \(x_{f}\) when the block comes to a complete stop.

The acceleration of a particle is given by \(a=-k s^{2}\) where \(a\) is in meters per second squared, \(k\) is a constant, and \(s\) is in meters. Determine the velocity of the particle as a function of its position \(s\). Evaluate your expression for \(s=5 \mathrm{m}\) if \(k=0.1 \mathrm{m}^{-1} \mathrm{s}^{-2}\) and the initial conditions at time \(t=0\) are \(s_{0}=3 \mathrm{m}\) and \\[ v_{0}=10 \mathrm{m} / \mathrm{s} \\]
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