A bomber is flying horizontally at an altitude of 3200 ft with a velocity of 400 ft/sec when it releases a bomb. A projectile is launched 5 seconds later from a cannon at a site facing the bomber and 5000 feet from the point where the bomb was released
Answers
If the bomb were dropped from a helocopter, the vertical position of the bomb would be the same as the vertical position from the bomber. People find that hard to believe, but if you watch a movie like Dr. Strangelove, you see it for yourself when Slim Pickens rides his payload (an atomic bomb) to earth.
So let's drop the bomb vertically from 3200 feet to 1600 feet and see how long it takes to get there. It will also give us how far the bomb moves horizontally.
Vi = 0 in the vertical direction
d = 3200 ft - 1600 feet = 1600 feet
a = 32 m/s^2
t = ?
d = Vi*t + 1/2 a*t^2
1600 ft = 0 + 1/2 * 32 ft/s^2 * t^2
1600 = 16 t^2 Divide by 16
100 = t^2 Take the square root of both sides
t = 10 sec.
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How far has the bomb gone horizontally? There is no horizontal acceleration so the formula you use (one of the very few places it can be used) is d = v*t
v = 400 feet horizontally. I can't emphasize this enough. 400 does not operate in the vertical.
t = 10 seconds
d = r*t = 400* 10 = 4000 feet horizontally.
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Too bad. This is where I have to stop. The bomb can only travel 4000 feet before it is intercepted. Perhaps your original height is incorrect. What follows is pretty good: I've never seen a problem like this and would like to continue, but can't. Maybe the original height was 32000 feet? 6 miles up is not impossible.