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Chapter 3: Problem 51

You are working as a consultant on a video game designing a bomb site for a World War I airplane. In this game, the plane you are flying is traveling horizontally at \(40.0 \mathrm{m} / \mathrm{s}\) at an altitude of $125 \mathrm{m}$ when it drops a bomb. (a) Determine how far horizontally from the target you should release the bomb. (b) What direction is the bomb moving just before it hits the target?

Short Answer

Expert verified
Answer: The horizontal distance to the target is approximately 202 meters, and the bomb is moving at an angle of approximately 51.2 degrees below the horizontal just before it hits the ground.

Step by step solution

01

(Step 1: List Known Values)

Known values are: - Horizontal velocity of the plane, \(v_x = 40 m/s\) - Altitude of the plane, \(h = 125 m\) - Acceleration due to gravity, \(g = 9.81 m/s^2\)
02

(Step 2: Calculate the Time of Flight)

First, we need to find out how long it takes for the bomb to reach the ground, which is also called time of flight. From the equation \(y = v_{0y}t - \frac{1}{2}gt^2\), we have: $$ h = 0 \cdot t - \frac{1}{2} \cdot g \cdot t^2 $$ Solving for \(t\): $$ t = \sqrt{\frac{2h}{g}} \text{, as } v_{0y} = 0 $$ Plugging in the known values: $$ t = \sqrt{\frac{2 \cdot 125}{9.81}} \approx 5.05 s $$ So, the time of flight for the bomb is \(t \approx 5.05 s\).
03

(Step 3: Calculate the Horizontal Distance)

Now, we can find the horizontal distance \(x\) by using the following equation: $$ x = v_x \cdot t $$ Plugging the known values: $$ x = 40 \cdot 5.05 \approx 202 m $$ Thus, the horizontal distance to the target is approximately \(202 m\).
04

(Step 4: Calculate the Final Vertical Velocity)

To find the final vertical velocity \(v_{y}\) just before the bomb hits the ground, we will use the equation: $$ v_{y} = v_{0y} - g \cdot t $$ Plugging the values: $$ v_{y} = 0 - 9.81 \cdot 5.05 \approx -49.6\ m/s $$
05

(Step 5: Calculate the Direction of the Bomb)

Now that we have both horizontal and vertical components of the velocity, we can calculate the angle \(\theta\) using the following formula: $$ \theta = \arctan{\frac{-v_{y}}{v_x}} $$ Plugging the values: $$ \theta = \arctan{\frac{49.6}{40}} \approx 51.2^\circ $$ The answer to (a) is that the bomb should be released approximately \(202 m\) horizontally from the target. For (b), the bomb is moving at an angle of approximately \(51.2^\circ\) below the horizontal just before it hits the target.

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