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A bomber is diving toward its target at an angle of 45ᶱ below the horizontal and at a speed of 320 m/s. When the bomber is 600 m above the ground, it releases its load, which then hits the target.

A) How long will the bomb take to reach the target?
B) What horizontal distance will the bomb travel?
C) With what speed will it strike the target?

User Sean Chase
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1 Answer

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Final answer:

The bomb will take a certain amount of time to reach the target, travel a certain horizontal distance, and strike the target at a certain speed. These values can be calculated using equations of projectile motion.

Step-by-step explanation:

The time it will take for the bomb to reach the target can be found using the vertical motion equation:

h = ut + (1/2)gt²

Where h is the height (600 m above the ground), u is the initial vertical velocity (component of the total velocity along the vertical direction), g is the acceleration due to gravity (9.8 m/s²), and t is the time taken.

Given that the angle of descent is 45 degrees below the horizontal, the vertical velocity can be found using the trigonometric relationship:

u = Vsinθ

Where V is the total velocity (320 m/s) and θ is the angle of descent (45 degrees).

By substituting the values into the equations, we can solve for t, which will give us the time taken for the bomb to reach the target.

Using the horizontal motion equation:

s = ut

Where s is the horizontal distance traveled, u is the initial horizontal velocity (component of the total velocity along the horizontal direction), and t is the time taken. Since the bomb is released horizontally, the initial horizontal velocity is equal to the total velocity.

By substituting the values into the equation, we can solve for s, which will give us the horizontal distance traveled by the bomb.

The speed at which the bomb strikes the target is equal to the total velocity, which is 320 m/s.

User Coree
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