Problem 10 - College Physics

Question

A military helicopter on a training mission is flying horizontally at a speed of 60.0\(\mathrm{m} / \mathrm{s}\) when it 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 the bomb's velocity just before it strikes the earth. (d) Draw graphs of the horizontal distance vs. time and the vertical distance vs. time for the bomb's motion. (e) If the velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground?

Step-by-Step Solution

Answer

(a) 7.82 s, (b) 469.2 m, (c) 60 m/s horizontally and 76.6 m/s vertically, (e) 469.2 m ahead.

Step 1: Determine Time to Reach the Ground

To find how long the bomb takes to reach the ground, we use the formula for vertical displacement due to gravity, \[ y = v_{i_y} t + \frac{1}{2} g t^2 \] where \( y = -300 \) m (as the bomb is falling downward), \( v_{i_y} = 0 \) (vertical initial velocity), and \( g = 9.8 \ \text{m/s}^2 \). Solving for \( t \):\[-300 = 0 + \frac{1}{2} (9.8) t^2 \]\[t^2 = \frac{-300}{-4.9} \]\[t^2 = 61.22 \]\[t = \sqrt{61.22} \approx 7.82 \ \text{s}\].

Step 2: Calculate Horizontal Distance

The horizontal distance traveled by the bomb can be found using the formula \[ x = v_{i_x} t \] where \( v_{i_x} = 60 \ \text{m/s} \) and \( t = 7.82 \ \text{s} \). Thus,\[ x = 60 \times 7.82 \approx 469.2 \ \text{m} \].

Step 3: Determine Components of Final Velocity

To find the horizontal component, note that it remains constant: \( v_{f_x} = v_{i_x} = 60 \ \text{m/s} \). The vertical component can be determined by \[ v_{f_y} = v_{i_y} + g t \] \[ v_{f_y} = 0 + 9.8 \times 7.82 \approx 76.6 \ \text{m/s} \].

Step 4: Horizontal and Vertical Distance Graphs

For horizontal distance vs. time, the graph is a straight line since the horizontal speed is constant. For vertical distance vs. time, the graph is a parabola opening upwards, because the only force acting is gravity (constant acceleration downwards).

Step 5: Helicopter's Position When the Bomb Hits

The helicopter continues at a constant speed of 60 m/s for 7.82 seconds. Therefore, \[ \text{distance} = 60 \times 7.82 \approx 469.2 \ \text{m}\], the helicopter will be directly above the bomb when it hits the ground.

Key Concepts

Vertical Displacement

Vertical displacement is the movement of an object in the vertical direction due to gravity. In this exercise, the bomb is initially at 300 meters above the ground, and it falls downward, which is why we denote the vertical displacement as negative 300 meters.
The formula used to calculate the time needed for an object to reach the ground under constant gravitational acceleration is:

• \( y = v_{i_y} t + \frac{1}{2} g t^2 \)

Here, \( y \) is the vertical displacement, \( v_{i_y} \) is the initial vertical velocity (0 m/s in this case as the bomb is only moving horizontally at the start), and \( g \) is the acceleration due to gravity, approximately 9.8 m/s².
By inserting these values and solving the equation, you find that the time it takes for the bomb to hit the ground is approximately 7.82 seconds. The negative sign for \( y \) simply indicates the downward direction.

Horizontal Distance

Horizontal distance refers to how far an object travels along the horizontal axis as it moves through the air. For projectiles, this distance is affected by the initial horizontal velocity and the time the object spends in the air.
The horizontal motion of a projectile can be calculated using the equation:

• \( x = v_{i_x} \, t \)

Where \( x \) is the horizontal distance, \( v_{i_x} \) is the initial horizontal velocity, and \( t \) is the time duration. In this problem, the initial horizontal velocity is 60 m/s, and the time calculated for the bomb to reach the ground is 7.82 seconds.
This results in a horizontal travel distance of approximately 469.2 meters. It's essential to notice that air resistance is ignored, so the horizontal velocity remains constant throughout the bomb's descent.

Velocity Components

The velocity of a projectile in motion can be split into two independent components: horizontal and vertical.

• The horizontal velocity component \( v_{i_x} \) remains constant throughout because there is no horizontal acceleration in the absence of air resistance. So, the horizontal component just before impact is the same as the initial horizontal velocity, 60 m/s.
• The vertical velocity component \( v_{f_y} \), however, changes over time due to the acceleration caused by gravity. The final vertical velocity can be calculated using:
  • \( v_{f_y} = v_{i_y} + g t \)
  Here, \( v_{i_y} \) is the initial vertical velocity (which is 0), and \( t \) is the time. Substituting known values gives a final vertical velocity of approximately 76.6 m/s at the moment the bomb impacts the ground.

This division into components is crucial for accurately analyzing the motion and determining the final velocity of a projectile.

Constant Acceleration

Constant acceleration is a key concept in projectile motion and describes motion that changes velocity at a uniform rate. In this exercise, gravity plays the role of a constant force, providing a steady acceleration downwards of about 9.8 m/s².
This consistent force affects only the vertical motion of the bomb. Since air resistance is disregarded, the horizontal motion experiences no acceleration and remains steady.
It's crucial to understand that:

• In vertical motion, the constant acceleration impacts how quickly an object speeds up or slows down over time, which is why the vertical velocity component increases steadily until the point of impact.
• In comparison, the horizontal motion is not subjected to any accelerative forces, thereby making it a straightforward uniform motion scenario.

By appreciating the role of constant acceleration, we can accurately predict how far, high, or fast an object will travel when subjected to consistent forces like gravity.