If an object is dropped, the distance it falls is given by D=1/2gt^2 where g is about 32 ft. If this is correct explain why..if not explain why and give correct method!
oh ya and the number after ^ is the exponet just to letcha kno
2007-10-29 18:30:07 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
wait how long would it take to fall at 1600feet???????????
2007-10-30 04:08:32 · update #1
This is correct except for one clarification. g is the constant of acceleration of gravity and is approximately 32 ft / sec². You just left off part of the units.
But the equation is correct. Assuming that an object is dropped with no initial velocity, the distance it has fallen after time t will be:
D(t) = ½gt²
D(0) = 0 ft.
D(1) = 16 ft.
D(2) = 64 ft.
D(3) = 144 ft.
etc.
2007-10-29 22:32:37 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
this is wrong for 2 reasons...
1) there is no mention of what speed (velocity) it was initially dropped with, it assumes that it was 0
2) there is no mention of what original distance it starts at. Right now, the equation says that this object was dropped from the reference frame of 0 and went down. this may be correct, but the question needs to specify this also...
The proper way to write it is:
D = Do + Vot - 1/2 gt^2
2007-10-29 18:37:36 · answer #2 · answered by sayamiam 6 · 0⤊ 0⤋