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Chapter 3: Problem 10

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?

Short Answer

Expert verified
(a) 7.82 s, (b) 469.2 m, (c) 60 m/s horizontal, 76.6 m/s vertical, (d) see steps, (e) 469.2 m away horizontally.

Step by step solution

01

Determine the time to reach the ground

To find the time it takes for the bomb to fall to the ground, use the equation of motion for vertical displacement: \[ y = v_{y0} t + \frac{1}{2} g t^2 \] The initial vertical velocity \(v_{y0}\) is 0 (since the bomb is dropped and not projected), \(y = 300\,\text{m}\), and \(g = 9.8\,\text{m/s}^2\) (acceleration due to gravity). Substitute the values:\[ 300 = 0 + \frac{1}{2} \times 9.8 \times t^2 \]Solve for \(t\):\[ 300 = 4.9 t^2 \]\[ t^2 = \frac{300}{4.9} \]\[ t^2 \approx 61.22 \]\[ t \approx \sqrt{61.22} \approx 7.82 \,\text{s} \]
02

Calculate horizontal distance travelled

The horizontal motion is uniform, as there are no horizontal forces acting on the bomb (ignoring air resistance). Calculate the horizontal distance using:\[ x = v_{x0} t \]where \(v_{x0} = 60 \,\text{m/s}\) and \(t = 7.82 \,\text{s}\).\[ x = 60 \times 7.82 \approx 469.2 \,\text{m} \]
03

Find the components of velocity when striking the ground

The horizontal component of velocity \(v_{x}\) remains 60 \,\text{m/s}\, constant. For the vertical component, use the equation:\[ v_{y} = v_{y0} + g t \]\[ v_{y} = 0 + 9.8 \times 7.82 \approx 76.6 \, \text{m/s} \]
04

Draw motion graphs

- **\(x-t\) graph**: A straight line with a positive slope, indicating constant horizontal velocity (slope = 60 m/s).- **\(y-t\) graph**: A curve that starts at 0, increasing its slope, representing increasing vertical velocity.- **\(v_{x}-t\) graph**: A horizontal line at 60 m/s, indicating constant horizontal velocity.- **\(v_{y}-t\) graph**: A straight line with a positive slope, starting from 0, representing linear increase in vertical velocity due to gravity.
05

Determine the helicopter's position when the bomb hits

Since the helicopter also travels at 60 \,\text{m/s}\, it covers the same horizontal distance as the bomb during the time \(t = 7.82 \,\text{s}\). Thus, it is horizontally \(469.2 \,\text{m}\) from the drop point.

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Key Concepts

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

Equations of Motion
Equations of motion are essential for understanding the trajectory of projectiles, such as a bomb dropped from a helicopter. These equations describe the relationship between an object's displacement, initial velocity, acceleration due to gravity, and time.

For vertical motion, when an object falls freely, the following equation is used:
  • Displacement: \( y = v_{y0} t + \frac{1}{2} g t^2 \)
  • Initial vertical velocity \( v_{y0} \) is often zero if the object is dropped.
  • Gravity \( g \) is approximately \( 9.8 \,\text{m/s}^2 \) on Earth.
For horizontal motion, where there is no acceleration because air resistance is ignored, the distance can be calculated by:
  • \( x = v_{x0} t \)
  • \( v_{x0} \) is the initial horizontal velocity, which remains constant.
These equations allow us to solve for variables such as time, horizontal distance, and final velocities before impact.
Vertical and Horizontal Components
Projectiles experience two independent motions: horizontal and vertical. Understanding the separation of these components is crucial for analyzing the entire trajectory.

**Horizontal Component**:
  • The horizontal velocity \( v_{x} \) remains constant, as there are no forces like air resistance acting horizontally.
  • In our example, \( v_{x0} = 60 \,\text{m/s} \).
**Vertical Component**:
  • The vertical motion is affected by gravity, causing acceleration downward.
  • Initially, if an object is simply dropped, \( v_{y0} = 0 \).
  • The vertical velocity builds as \( v_{y} = v_{y0} + gt \).
This separation simplifies the calculations, as horizontal motion involves constant speed, while vertical motion involves constantly changing speed due to gravity.
Velocity-Time Graphs
Velocity-time graphs provide a visual representation of how velocity changes over time for both horizontal and vertical components of motion. They are invaluable for visual learners and offer insight at a glance.

**Horizontal Velocity-Time Graph (\( v_{x}-t \))**:
  • A horizontal line indicating constant velocity (e.g., 60 m/s).
**Vertical Velocity-Time Graph (\( v_{y}-t \))**:
  • A line with a positive slope starting from zero, showing velocity increase due to gravity.
  • Ending velocity can be determined by the final point of the line.
These graphs help in quickly identifying the behavior of the projectile's speed over its flight, reinforcing the separation of motion components.
Free Fall
Free fall describes the motion of objects under the sole influence of gravity, with no other forces, such as air resistance, acting. It is a special case of projectile motion, where the vertical component of motion is emphasized.

Key Characteristics of Free Fall:
  • Initial vertical velocity \( v_{y0} \) is often zero if an object is dropped rather than thrown.
  • Only gravity acts, causing a constant acceleration of \( 9.8 \,\text{m/s}^2 \).
  • Time in free fall affects both velocity and displacement vertically.
In the context of the helicopter bomb example, the progressive increase of vertical velocity due to gravity alone exemplifies free fall, contrasting the constant horizontal motion. Free fall calculations can predict impact time and final impact velocity, aiding accurate predictions of landing.

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Most popular questions from this chapter

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?

A 400.0 -m-wide river flows from west to east at 30.0 \(\mathrm{m} / \mathrm{min}\) . Your boat moves at 100.0 \(\mathrm{m} / \mathrm{min}\) relative to the water no matter which direction you point it. To cross this river, you start from a dock at point \(A\) on the south bank. There is a boat landing directly opposite at point \(B\) on the north bank, and also one at point \(C, 75.0 \mathrm{m}\) downstream from \(B(\text { Fig. } 3.53)\) . (a) Where on the north shore will you land if you point your boat perpendicular to the water current, and what distance will you have traveled? (b) If you initially aim your boat directly toward point \(C\) and do not change that bearing relative to the shore, where on the north shore will you land? (c) To reach point \(C :(i)\) at what bearing must you aim your boat, (ii) how long will it take to cross the river, (iii) what distance do you travel, and (iv) and what is the speed of your boat as measured by an observer standing on the river bank?

Two students are canoeing on a river. While heading effort? upstream, they accidentally drop an emply bottle oveabound. They then continue paddling for 60 minutes, reaching a point 20 \(\mathrm{km}\) farther upstream. At this point they realize that the bottle is missing and, driven by ecological awareness, they tum around and head downstram. They catch up with and retrieve the bottle (which has been moving along with the current) 5.0 \(\mathrm{km}\) down- stream from the tum-around point. (a) Assuming a constant paddling effort throughout, how fast is the river flowing? (b) What would the canoe speed in a still lake be for the same paddling effort?

Kicking a Field Goal. In U.S. football, after a touchdown the team has the opportunity to earn one more point by kicking the ball over the bar between the goal posts. The bar is 10.0 \(\mathrm{ft}\) above the ground, and the ball is kicked from ground level, 36.0 \(\mathrm{ft}\) horzontally from the bar (Fig. 3.48\() .\) Football regulations are stated in English units, but convert to SI units for this problem. (a) There is a minimum angle above the ground such that if the ball is launched below this angle, it can never clear the bar, no matter how fast it is kicked. What is this angle? (b) If the ball is kicked at \(45.0^{\circ}\) above the horizontal, what must its initial speed be if it to just clear the bar? Express your answer in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{km} / \mathrm{h}\) .

On the Flying Trapeze. new circus act is called the Texas Tumblers. Lovely Mary Belle swings from a trapeze, projects herself at an angle of \(53^{\circ},\) and is supposed to be canght by Joe Bob, whose hands are 6.1 \(\mathrm{m}\) above and 8.2 \(\mathrm{m}\) horizontally from her launch point (Fig. 3.50 ) You can ignore air resistance. (a) What initial speed \(v_{0}\) must Mary Belle have just to reach Joe Bob? (b) For the initial speed calculated in part (a), what are the magnitude and direction of her velocity when Mary Belle reaches Joe Bob? (c) Assuming that Mary Belle has the initial speed calculated in part (a), draw \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs showing the motion of both tumblers. Your graphs should show the motion up until the point where Mary Belle reaches Joe Bob. (d) The night of their debut performance, Joe Bob misses her completely as she flies past. How far horizontally does Mary Belle travel, from her initial launch point, before landing in the safety net 8.6 \(\mathrm{m}\) below her starting point?
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