31st August 2005 - 11:06 PM
An airplane is flying at an altitude of 8000 meters at a speed of 900 km/hr. At what distance ahead of a target must it drop its relief cargo? If the plane turns around at the moment the cargo is released, how far from the target will it be at the time the cargo hits the target?
Envisoin the plane as a static location in space. In other words think of the bomb or the cargo as a ball and the plane as a cliff.
So let me reword the question without harming the actual numbers and facts:
A ball falls from a cliff 8000 meters high with an initial horizontal velocity of 900km/hr. It hits the sea at a farther distance of Y.
A) find Y.
B) if a person starts to move away from the tip of the cliff at 900km/hr(say Superman or Bush) how far is he from the tip, by the time the ball he shot hits the sear floor.
Ok so now the question is more friendly and u can right away see that this is in fact a PROJECTILE problem.
Before I answer the question, I can give you a hint. A physics professor or a text book knows that the average physics student knows the formula and probably solves the samples. What students however usually miss is that the questions are always very easy but they have to crack the shell of the question and put behind the literature to get to the hard core physics. Once you do that, the question is just a matter of the substitution hence the old saying "physics is just about substituing formuals".
Ok now lets do part A:"
we know that Vx or velocity in horizonal direction of the ball is Vx = 900 km/hr assuming that the plane merely releases the ball and DOES NOT fire it. We also konw that the vertical velocity of the ball or Vy = 0 m/s because the ball and the plane are ideally paralle to the tanget of the curvature of the earth. Ok so lets write down yet again our data:
Vx = 900 km /hr = 900,000 m / 3600 sec. = 250 m/s
(Dont forget the conversion)
Vy1 = 0.0 m/s
g = 9.80 m/s^2 (assuming we look at this as a classical physics problem and not one involving relativity and contraction of time and space)
Y = distance from the the tip of the cliff to the place of impact on the sea (In other words the horizontal distance from the point of release of the bomb or the package and the point of intersection with water)
H = height of the cliff = 8,000 meters
Now lets review PROJECTILE forumals.
First of all this is a flat projectile and we are experiencing a quarter of a circle kind of motion.
the followings are the formuals we know of projectile:
Vx = Y / t [For Horizontal Motion] (note the horizontal velocity is constant)
H = Vy1t + 1/2 g (t)^2 [For Vertical Motion] (the vertical velocity is increasing)
Usually when answering Projectile problems, u use a fusion of the two formulas. U notice that the t or time is the same for both because they are simultanously occuring.
Now, lets see just plug in all that we have:
(1) 250 = Y / t
(2) 8000 = 0*t + 1/2 * 9.80 * t^2
so if we find t we can find Y but first we have to solve (2):
8000 = 1/2 * 9.80 * t^2 SO t = 40 seconds
Now, we find Y = 250 * 40 = 10,000 meters
So we find the distance from which the ball would hit the ground, and that point is almost 10 km
Part B) now that u have the time or t = 40 seconds. Since the plane turns on a dime meaning it wont waste any time on turning it is then moving at a constant speed of 250 m/s and in 40 seconds how far is it?
250 = d / t = d / 40
d = 250 * 40 = 10000 or 10 km
An interesting fact: Messerschmitt Me planes had a special jet engine that would use air like a projectile and aid in fuel efficiency.
Also when US took care of japan, they had to first calculate the height of the release and the speed of the plane to make sure that the blast radius of the atom did not catch their own planes. So even simple physics has some good uses.
Dr. David, a NASA scientist proved his innocence in a case of fire in near by stable by use of projectile. David used to experiment with jets and experimental science projects. Accidentaly a flair from an army base nearby caught a house on fire and he was falsly accused for it and he repudiated the claim by explaning that considering the max velocity and angle of his projectile and the wind speed, it is impossible for him to have reached the house...another beautiful application of physics.