91Ó°ÊÓ

Your grandfather is copilot of a bomber, flying horizontally over level terrain, with a speed of 275 m/s relative to the ground, at an altitude of 3000 m. (a) The bombardier releases one bomb. How far will it travel horizontally between its release and its impact on the ground? Neglect the effects of air resistance. (b) Firing from the people on the ground suddenly incapacitates the bombardier before he can call, "Bombs away!" Consequently, the pilot maintains the plane's original course, altitude, and speed through a storm of flak. Where will the plane be when the bomb hits the ground? (c) The plane has a telescopic bomb sight set so that the bomb hits the target seen in the sight at the time of release. At what angle from the vertical was the bomb sight set?

Step by step solution

01

Determine the time of flight of the bomb

To find the time the bomb takes to hit the ground, we'll use the second equation of motion i.e. \(d = ut + 0.5gt^2\), where d is distance (equal to the altitude 3000 m), u is initial vertical velocity (equal to 0, since the bomb is dropped and not thrown vertically), g is acceleration due to gravity (approximately 9.8 m/s^2), and t is time taken. Solving for t, we get \(t = \sqrt{{2d}/{g}}\).

02

Compute the horizontal distance travelled by the bomb

The horizontal distance the bomb will travel can be calculated using the formula \(d = vt\) where v is the horizontal velocity (equal to the speed of the plane, 275 m/s) and t is time (calculated in the previous step).

03

Determine the plane's position when the bomb hits

The plane will be in the same position as the bomb when it hits the ground, since the horizontal speed of the bomb is equal to the horizontal speed of the plane and they were released at the same time.

04

Calculate the angle of the bomb sight

The bomb sight would have been set so that the bomb hits the target seen at the time of release. This means the bomber would be looking directly down the vertical. Hence, the angle from the vertical is 0 degrees.

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Key Concepts

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

Horizontal Distance Calculation

In projectile motion, understanding how to calculate the horizontal distance traveled by an object is crucial. Here, the horizontal distance, also known as the range, can be found using the formula for horizontal motion.

Horizontal motion assumes that there is no acceleration in the horizontal direction once an object is in motion. This is because, according to the problem, air resistance is neglected. The only force acting on the object is gravity, which affects the vertical component of motion. To find the horizontal distance, we'll use the formula:

• \( d = v \times t \)

In this scenario:

• \( v \) represents the horizontal velocity of the plane, which is constant and given as 275 m/s.
• \( t \) is the time of flight, which we have to calculate first using vertical motion principles.

Once you know the time of flight, simply multiply it by the horizontal velocity to get the horizontal distance. This straightforward multiplication stems from having a constant horizontal speed.

Time of Flight

To determine how long an object is in the air, known as the time of flight, we must consider the vertical motion. The time a projectile takes to hit the ground depends on the height from which it is dropped and gravity, assuming vertical launch velocity is zero, like in our example.

Let's use the second equation of motion to find the time of flight:

• \( d = ut + 0.5gt^2 \)

For this problem:

• \( d = 3000 \) m, representing the altitude.
• \( u = 0 \) m/s, since the bomb is released and not thrown.
• \( g = 9.8 \) m/s² is the acceleration due to gravity.

Substitute these into the formula to solve for \( t \):

• \( 3000 = 0 + 0.5 \times 9.8 \times t^2 \)
• \( t = \sqrt{{2 \times 3000}/{9.8}} \)

This equation simplifies to provide the time it takes for the bomb to reach the ground. The square root is taken to ensure we obtain a positive value for time, which is physically meaningful.

Bomb Sight Angle

The bomb sight angle is key in aiming accurately in projectile motion scenarios. In simple terms, this angle determines how the bomb is directed towards a target.

In our exercise, the plane's bomb sight is configured so that the target is lined up vertically when the bomb is released. This suggests that:

• The bomb sight angle from the vertical should be zero degrees.
• This implies the bombardier would look straight down, parallel to gravity, ensuring the bomb lands directly on the target below as seen at the moment of release.

This configuration assumes perfect conditions where factors such as wind aren't considered, making it easier to predict the bomb's path. In real-world scenarios, additional adjustments might be necessary to account for such variables. However, in this ideal case, knowing the sight angle as zero underlines the precision and simplicity of the task at hand without external forces acting upon the bomb’s trajectory.