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Chapter 5: Problem 4

A helicopter is flying at \(200 \mathrm{~m}\) and flying at \(25 \mathrm{~m} \mathrm{~s}^{-1}\) at an angle \(37^{\circ}\) above the horizontal when a package is dropped from it. The distance of the point from point \(O\) where the package lands is (1) \(80 \mathrm{~m}\) (2) \(100 \mathrm{~m}\) (3) \(200 \mathrm{~m}\) (4) \(160 \mathrm{~m}\)

Short Answer

Expert verified
The package lands approximately 100 meters from point O.

Step by step solution

01

Determine Initial Velocity Components

The helicopter is flying at an initial velocity of \(25 \; \mathrm{m/s}\) at an angle \(37^{\circ}\) above the horizontal. Break this velocity into horizontal and vertical components:- Horizontal component \(v_{x} = 25 \cdot \cos(37^{\circ})\)- Vertical component \(v_{y} = 25 \cdot \sin(37^{\circ})\)Using \(\cos(37^{\circ}) \approx 0.8\) and \(\sin(37^{\circ}) \approx 0.6\):- \(v_{x} = 25 \times 0.8 = 20 \; \mathrm{m/s}\)- \(v_{y} = 25 \times 0.6 = 15 \; \mathrm{m/s}\)
02

Calculate Time of Flight

The initial vertical velocity component \(v_{y} = 15 \; \mathrm{m/s}\) and the package is dropped from a height of \(200 \; \mathrm{m}\). To find the time of flight \(t\), use the equation of motion:\[ h = v_{y} \cdot t + \frac{1}{2} g \cdot t^2\]Substituting the values:\[200 = 15 \cdot t + \frac{1}{2} \times 9.8 \times t^2\]Rewriting, we have:\[4.9t^2 + 15t - 200 = 0\]Solve this quadratic equation to find \(t\). Use the quadratic formula:\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Where \(a = 4.9\), \(b = 15\), \(c = -200\). Calculate \(t\).
03

Solve the Quadratic Equation for Time

Using the quadratic formula, solve:\[t = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 4.9 \cdot (-200)}}{2 \cdot 4.9}\]Calculate the discriminant:\[b^2 - 4ac = 15^2 + 4 \times 4.9 \times 200 = 225 + 3920 = 4145\]Thus,\[t = \frac{-15 \pm \sqrt{4145}}{9.8}\]Approximate \(\sqrt{4145} \approx 64.38\).Find the positive root:\[t = \frac{-15 + 64.38}{9.8} \approx \frac{49.38}{9.8} \approx 5.04 \, \mathrm{s}\]
04

Calculate Horizontal Distance

With the time of flight \(t \approx 5.04 \, \mathrm{s}\) and horizontal velocity \(v_{x} = 20 \, \mathrm{m/s}\), calculate the horizontal distance \(d\):\[d = v_{x} \cdot t = 20 \, \mathrm{m/s} \times 5.04 \, \mathrm{s} \approx 100.8 \, \mathrm{m}\]Thus, the distance from point \(O\) where the package lands is approximately 100 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 an object is launched into the air, it's crucial to decompose its initial velocity into horizontal and vertical components. This is because these components act independently of each other during projectile motion. Here, the helicopter's total velocity is given as \(25 \, \text{m/s}\) at an angle of \(37^{\circ}\) above the horizontal.

To find the horizontal component \(v_{x}\), use the cosine of the angle:
  • \(v_{x} = 25 \cdot \cos(37^{\circ})\)
  • Approximation: \(\cos(37^{\circ}) \approx 0.8\)
  • Calculation: \(v_{x} = 25 \times 0.8 = 20 \, \text{m/s}\)
Similarly, calculate the vertical component \(v_{y}\) using the sine of the angle:
  • \(v_{y} = 25 \cdot \sin(37^{\circ})\)
  • Approximation: \(\sin(37^{\circ}) \approx 0.6\)
  • Calculation: \(v_{y} = 25 \times 0.6 = 15 \, \text{m/s}\)
This decomposition allows us to independently analyze the horizontal and vertical motions.
Time of Flight
The time of flight refers to how long the package spends in the air before it hits the ground. In projectile motion, the vertical motion primarily determines the time of flight. The equation of motion for vertical displacement is used here:
  • \(h = v_{y} \cdot t + \frac{1}{2} g \cdot t^2\)
Given:
  • Initial height \(h = 200 \, \text{m}\)
  • Initial vertical velocity \(v_{y} = 15 \, \text{m/s}\)
  • Gravitational acceleration \(g = 9.8 \, \text{m/s}^2\)
Substitute into the equation:
  • \(200 = 15t + \frac{1}{2} \times 9.8 \times t^2\)
  • Reorganize to form a quadratic equation: \(4.9t^2 + 15t - 200 = 0\)
Solve this to find the flight time \(t\). This process involves using the quadratic equation, a standard form to solve for unknowns where the polynomial is second degree.
Quadratic Equation
Solving the time of flight involves a quadratic equation of the form \(ax^2 + bx + c = 0\). Applying the quadratic formula:
  • \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Use the coefficients from our equation:
  • \(a = 4.9\), \(b = 15\), \(c = -200\)
  • Discriminant: \(b^2 - 4ac\)
  • Calculate: \(15^2 - 4 \times 4.9 \times (-200) = 225 + 3920 = 4145\)
With the discriminant calculated, obtain the square root: \(\sqrt{4145} \approx 64.38\).
Validating the flight duration involves choosing the positive root, leading to the result:
  • \(t = \frac{-15 + 64.38}{9.8} \approx 5.04 \, \text{s}\)
Horizontal Distance Calculation
Horizontal distance refers to how far the package travels horizontally while in the air. With the obtained time of flight \(t\) and the horizontal velocity component \(v_x\), calculate this distance. The formula is straightforward:
  • \(d = v_{x} \cdot t\)
Given:
  • \(v_{x} = 20 \, \text{m/s}\)
  • \(t \approx 5.04 \, \text{s}\)
Calculate the distance:
  • \(d = 20 \times 5.04 \approx 100.8 \text{m}\)
This result shows how far the package lands from its original point of release, demonstrating the principles of projectile motion in action.

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Most popular questions from this chapter

A helicopter is flying at \(200 \mathrm{~m}\) and flying at \(25 \mathrm{~m} \mathrm{~s}^{-1}\) at an angle \(37^{\circ}\) above the horizontal when a package is dropped from it. If the helicopter flies at constant velocity, find the \(x\) and coordinates of the location of the helicopter when the \(v\) package lands. (1) \(160 \mathrm{~m}, 320 \mathrm{~m}\) (2) \(100 \mathrm{~m}, 200 \mathrm{~m}\) (3) \(200 \mathrm{~m}, 400 \mathrm{~m}\) (4) \(50 \mathrm{~m}, 100 \mathrm{~m}\)

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