Physics Problem HELP! please?
A stone is dropped down a well and 5 sec later the sound of the splash is heard, if the velocity of sound is 1120 ft/sec, what is the depth of the well?
A ball is dropped from the top of a tower 80 ft height at the same instant that a second ball is thrown upward from the ground with an initial velocity of 40 ft/sec. when and where do they pass and with what relative velocity?
please help:D
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The stone and the echo travel the same distance s.
The stone does it in t seconds, the echo does it in 5 - t seconds.
s = ½at² = 16.1ft/s² * t² ← for the stone
s = v(5 - t) = 1120ft/s * (5s - t) = 5600ft - 1120ft/s * t ← for the echo
These are equal.
16.1ft/s² * t² = 5600ft - 1120ft/s * t
quadratic in t; solutions at
t = -74.2s, 4.68s
Stone takes 4.68s to traverse the distance, echo takes 0.32s
Well depth is s = vt = 1120ft/s * 0.32s ≈ 360 ft
first ball: s = So + ½at² = 80ft - 16.1ft/s² * t²
second ball: S = Vo*t + ½at² = 40ft/s * t - 16.1ft/s² * t²
When are they equal? s = S, or
80 - 16.1t² = 40t - 16.1t²
80 = 40t
t = 2s
Check:
s = 80ft - 16.1ft/s² * (2s)² = 80ft - 64.4ft = 15.6 ft
S = 40ft/s * 2s - 16.1ft/s² * (2s)² = 80ft - 64.4ft = 15.6 ft √ is where they pass.
first: v = at = -32.2ft/s² * 2s = -64.4 ft/s
second: V = Vo + at = 40ft/s - 32.2ft/s² * 2s = -24.4 ft/s
So the dropped stone passes the thrown stone with a relative velocity of 40 ft/s.