A rock is dropped down a well. It takes 4 seconds to hear the splash approximately how deep is the well, to the nearest foot?
hint: h = -16.1t^2 + Vot
2007-07-17 17:55:47 · 5 answers · asked by wallst 1 in Science & Mathematics ➔ Mathematics
hint: Vo = 0 as it was dropped, not thrown.
If you want to blow your teacher away, make a correction for the speed of sound.
2007-07-17 17:59:48 · answer #1 · answered by jcsuperstar714 4 · 0⤊ 0⤋
Simple. Substitute for t = 4 secs.
h = -16.1 t^2 + Vot ( s = ut + 1/2 at^2 is the formula).
h = - 16.1 x 4 x 4 + 0 (since there is no initial velocity)
= 257.6 ft (the negative sign indicates depth).
We are ignoring the time taken for the sound to come back to our ears in the above calculation. The depth is actually less since time a fraction of a second to come back.
= 258 ft. to the nearest feet.
2007-07-18 02:45:37 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋
Some hint. Why don't you just substitute 4 for t in this equation. Vot=0, since there is no throwing of the rock downwards.
2007-07-18 01:00:58 · answer #3 · answered by cattbarf 7 · 0⤊ 0⤋
take t = 4 and Vo = 0 since the rock is dropped from rest and not thrown.
h = 16.1(4)^2
h = 257.6 ft
2007-07-18 01:03:23 · answer #4 · answered by Southpaw 5 · 0⤊ 0⤋
Distance traveled by the rock is the depth of well:
d = u*t + g * t^2; g = 9.8, u = 0
d = 9.8 * 16 meter
Find it out
2007-07-18 01:07:18 · answer #5 · answered by harshadanywhere 3 · 0⤊ 0⤋