I think you need more info for this problem. More specifically where the target is located.
Jameson
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.
thanks in advance.
Here is one way.Originally Posted by Aminhas18
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.
Hey, I missed that last night.Originally Posted by Aminhas18
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.