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Illustration of a military scene, showing a level fight bomber in action. In the sky, the bomber is rendered at high speeds indicating it's about 300 ft/s. A bomb has just been released from the aircraft which itself is at an elevation of 6400 ft. Below the aircraft, there's a landscape of open fields and mountains, portraying the vast distance the bomb must travel. Still, no objects, towns, or people are experienced in the target area. Also, note that the image includes no text.

A level fight bomber flying at 300 ft/s releases a bomb at an elevation of 6400 ft. Approximately how long before the bomb strikes the earth?

Only the vertical component counts for the time for descent.

Assuming no air resistance, we have from kinematics,
S=ut+(1/2)at^2
where
S=distance = 6400 ft
u=initial velocity (=0 ft/s vertical in this case)
a=acceleration due to gravity, g=32.2 ft/s²

Substitute values,
6400=(1/2)(32.2)t²

Solve for t.

T=20s

what is the formula

Well, I like to think that bombs have a sense of adventure, so they might take a little detour before striking the earth. But in all seriousness, to calculate the time it takes for the bomb to strike the earth, we need to use the formula: time = distance / speed.

Given that the bomber is flying at a speed of 300 ft/s and the bomb was released at an elevation of 6400 ft, the bomb needs to cover a distance of 6400 ft before hitting the earth.

Using the formula, time = distance / speed, we get time = 6400 ft / 300 ft/s = 21.33 seconds.

So, approximately 21.33 seconds before the bomb strikes the earth. Just enough time to appreciate the view, I suppose!

To determine how long it takes for the bomb to strike the earth, we need to calculate the time it would take for the bomb to fall from the release point to the ground.

We can use the kinematic equation for vertical motion:

y = y0 + v0*t - (1/2) * g * t^2

where:
y = vertical position (in this case, the height)
y0 = initial vertical position (launch point)
v0 = initial vertical velocity (velocity of the bomber)
g = acceleration due to gravity (approximately 32 ft/s^2)
t = time

From the information given, we have:
y0 = 6400 ft
v0 = 0 ft/s (as the bomb is released)
g = 32 ft/s^2

Plugging these values into the equation, we have:
y = 6400 + 0 * t - (1/2) * 32 * t^2

Simplifying the equation, we get:
y = 6400 - 16t^2

To find the time it takes for the bomb to hit the ground, we set y = 0 and solve for t:

0 = 6400 - 16t^2
16t^2 = 6400
t^2 = 400
t = √400
t = 20 seconds

Therefore, it would take approximately 20 seconds for the bomb to strike the earth after being released from the bomber.