Question 263100
One formula we can use is
(i) {{{h(t) = -9.8t^2 + V(sub 0) t + S(sub 0)}}}
we know that v(sub0) = 0
and if t= 1 then h(1) = 2
so, w get
2 = -9.8(1)^2 + S(sub0)
S(sub0) = 11.8
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Our equation becomes
(ii) {{{h(t) = -9.8t^2 + 11.8}}}
take a derivative to get speed as
(ii) s(t) = -19.6t
take another derivative to get acceleration as
(iii) a(t) = -19.6
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in the second second, we get
h(2) = -9.8*2^2 + 11.8
h(2) = -27.4 or 27.4 feet down hill.
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how fast at end of second second sounds like average speed or
average = (h(2)-h(1)) / (2-1)
or
average = (-27.4-2)/1 = -29.4
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how fast at end of first second sounds like average speed or
average = (h(1)-h(0)) / (1-0)
or
average = (2-11.8)/1 = -8.2
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