1. ## [SOLVED] Projectiles.

Hello,

I would greatly appreciate some help to figure out this question.

A bomber is flying at an altitude of 30,000 feet at a speed of 540 miles per hour. When should the bomb be released for it to hit the target?( Give your answer in terms of the angle of depression from the plane to the target.) What is the speed of the bomb at the time of impact.

2. I think you need more info for this problem. More specifically where the target is located.

Jameson

3. Actually you don't need more info, you just answer the question in reference to the target location something like 500 meters before flying directly over the target you drop the load. I'mn not sure what to make of the angle of depression though.

4. Originally Posted by MathGuru
Actually you don't need more info, you just answer the question in reference to the target location something like 500 meters before flying directly over the target you drop the load. I'mn not sure what to make of the angle of depression though.
The angle of depression is something that can be measured from
the bomber. It is essential the angle whose tangent is equal to
the bomber's height divided by the horizontal range to target at bomb
release.

RonL

5. Originally Posted by Aminhas18
Hello,
A bomber is flying at an altitude of 30,000 feet at a speed of 540 miles per hour. When should the bomb be released for it to hit the target?( Give your answer in terms of the angle of depression from the plane to the target.) What is the speed of the bomb at the time of impact.
Here is one way.

We disregard wind resistance to the bomb. The bomber is assumed as flying horizontally, so the bomb is "flying" horizontally also at time of release.

The horizontal distance to be covered by the bomb to hit the target is, say, D.
D = (540mph)*t -----since no horizontal external force "fights" the horizontal 540mph speed of the bomb.

The vertical distance to be covered by the bomb to hit the target is 30,000 ft.
h = Vo*t -(g/2)t^2 ------formula for vertical distance by a projectile.
Here, Vo, or initial vertical speed, is zero because the bomb is "dropped" only. Like the bomb is in free fall only.
And, since the vertical distance is downwards, the "-g" should be "+g" here. Or, the acceleration due to gravity should be positive here.
So,
30,000 = 0*t +(32/2)t^2
t = sqrt(30,000/16) = 43.3 sec
In hours, t = (43.3sec)*(1min/60sec)*(1hr/60min) = 43.3 /3600 = 0.012028hr

Hence, horizontal distance D = (540mph)(0.012028hr) = 6.495 miles.
In feet, D = (6.495mi)*(5280ft /1mi) = (6.495)(5280) = 34,293.6 ft.

Then, to get the angle of depression, theta, for the bomb to be released,
tan(theta) = (30,000ft) /(34,293.6ft) = 0.874799
theta = arctan(0.874799) = 41.1794 degrees = 41deg, 10min, 46sec

Therefore, the bomb should be released when the angle of depression to the target is 41deg, 10min, 46sec. -------answer.

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Of course, that is not very precise, even if the angle is in seconds of an angle already. The cumulative rounding off affected the precise angle.
The answer is just to show how it is solved.

6. Originally Posted by Aminhas18
Hello,

I would greatly appreciate some help to figure out this question.

A bomber is flying at an altitude of 30,000 feet at a speed of 540 miles per hour........ What is the speed of the bomb at the time of impact.
Hey, I missed that last night.

At impact, the speed or velocity of the bomb has vertical and horizontal components.

Horizontal component is 540 mph.

Vertical component is Vo +gt. Or, 0 +32(43.3) = 1385.6 ft/sec.
In mph, that is (1385.6 ft/sec)*[1mi /5280ft]*[3600sec /1hr] = (1385.6 *1*3600)/(5280*1) = 944.73 mph

So, the resultant velocity at impact is, using Pythagorean theorem,
sqrt[(540)^2 +(944.73)^2] = 1088.2 mph ------answer.

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# a bomber flying horizontally at a height of 500 with a velocity of 450 kmphcas hime to hit the target.find at what distance from the target,he should telease the bomb in order to hit the target

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