A package is dropped from a passing plane(moving in the horizantal direction) and produces a projectile motion as it is falling to the ground. If it takes the package only 49.92 seconds until it hits the ground, how far above the ground was the plane when the package was released?
2006-09-20 16:46:29 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Must be basic physics, where wind resistance is ignored. In that case, the time it takes to fall to the ground is the same as if it had been dropped from a fixed location.
therefore you can apply this basic equation, h = 1/2 g t^2
w/ g = gravitational constant, and t = time = 49.92sec
then, h = 1/2 (32.2 ft/sec^2) * (49.92sec)^2 = 40,121 ft
Wow. Falls fast!
2006-09-24 14:12:35 · answer #1 · answered by 2Horses 2 · 0⤊ 0⤋
If one makes the obvious assumption that the package was not pushed either upward or downward when released, and that air resistance can be ignored (which is a bit of a stretch if you are dropping the package from an airplane), then the standard gravitational equation
s = 0.5 g t^2 applies, and the answer can easily be figured with a four-banger.
2006-09-20 16:54:45 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Height= 1/2 g times time^2= 12460 metres.
2006-09-20 17:43:00 · answer #3 · answered by zee_prime 6 · 0⤊ 0⤋
39872.1 feet
h = (1/2)at^2 = 16 t^2 = 16* 49.92^2 = 39872.1
2006-09-20 16:56:27 · answer #4 · answered by nospamcwt 5 · 0⤊ 0⤋
need more info
2006-09-20 16:48:13 · answer #5 · answered by cosmic_convoy 3 · 0⤊ 1⤋