A rock is dropped off the edge of a cliff and its distance s (in feet) from the
ID: 2856327 • Letter: A
Question
A rock is dropped off the edge of a cliff and its distance s (in feet) from the top of the cliff after t seconds is s(t) = 16t^2. Assume the distance from the top of the cliff to the water below is 576 ft. Answer part (a) and (b). When will the rock strike the water? The rock will strike the water at seconds. (Type an integer or a simplified fraction.) Make a table of average velocities and approximate the velocity at which the rock strikes the water. The rock strikes the water at approximately ft/sec. (Round to the nearest integer as needed.)Explanation / Answer
a)rock will hit the bottom when s(t)=576
16t2=576
t2=576/16
t=24/4
t=6 seconds
after 6 seconds rock hits the bottom
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b)average velocity (5,6)
=[s(6)-s(5)]/(6-5)
=[16*62-16*52]/(6-5)
=176ft/s
average velocity (5.5,6)
=[s(6)-s(5.5)]/(6-5.5)
=[16*62-16*5.52]/(6-5.5)
=184ft/s
average velocity (5.9,6)
=[s(6)-s(5.9)]/(6-5.9)
=[16*62-16*5.92]/(6-5.9)
=190.4ft/s
average velocity (5.99,6)
=[s(6)-s(5.99)]/(6-5.99)
=[16*62-16*5.992]/(6-5.99)
=191.84ft/s
average velocity (5.999,6)
=[s(6)-s(5.999)]/(6-5.999)
=[16*62-16*5.9992]/(6-5.999)
=191.984ft/s
rock strikes the surface at approximately 192ft/s
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