Use the position function $s\left(t\right)=-16{t}^{2}+{v}_{0}t+{s}_{0}$ for free-falling objects. A silver dollar is dropped from the top of a building that is 1362 feet tall. (a) Determine the position and velocity functions for the coin. (b) Determine the average velocity on the interval [1, 2]. (c) Find the instantaneous velocities when . (d) Find the time required for the coin to reach ground level. (e) Find the velocity of the coin at impact.
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godsrvnt0706

Step 1
a) The position function:
in order to determine this equation we have to determine :
the coin is dropped with no initial velocity (free falling object) so we conclude that ${v}_{0}=0$.
When $t=0$ have $s\left(0\right)={s}_{0}=1362$ (because the coin was dropped from the top of the building with a height $=1362$)
$s\left(t\right)=-16{t}^{2}+{v}_{0}t+{s}_{0}$
${v}_{0}$ is the initial velocity (when $t=0$)
${s}_{0}$ is the initial position (when $t=0$)
(these concepts are from physics!)
we conclude that: the position function $=s\left(t\right)=-16{t}^{2}+1362$
Step 2
a) the velocity function $=\frac{d\left(s\left(t\right)\right)}{dt}$
$v\left(t\right)=\frac{d\left(s\left(t\right)\right)}{dt}=-32t$
Step 3
b) the average velocity $=\frac{s\left(2\right)-s\left(1\right)}{2-1}=s\left(2\right)-s\left(1\right)$
the average velocity $=s\left(2\right)-s\left(1\right)=-16×4+1362-\left(-16×1+1362\right)=-48$
Step 4
c) the instantaneous velocity at 2 is v(2)
the instantaneous velocity at 1 is v(1)
$v\left(2\right)=-32×2=-64$
$v\left(1\right)=-32$
Step 5
d) time required to the coin to reach the ground:
when coin reach the ground we have $s\left(t\right)=0$
so we must solve this equation to determine the time
$s\left(t\right)=0$
$⇒-16{t}^{2}+1362=0$
$⇒-16{t}^{2}=-1362$
$⇒{t}^{2}=\frac{1362}{16}=85.125$
$⇒t=\sqrt{85.125}\approx 9.23$
Step 6
e) plug the time when the coin reaches the ground into the velocity equation: $v\left(t\right)=-32t$
$v\left(\sqrt{85.125}\right)\approx v\left(9.23\right)=-32×9.23=-295.36$
a) $s\left(t\right)=-16{t}^{2}+1362,v\left(t\right)=-32t$
b) the average velocity $=-48$
c) $v\left(1\right)=-32,v\left(2\right)=-64$
d) $t\approx 9.23$
e) $t\approx -295.36$

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Elaine Verrett

Good grief, feet and inches and yards. That is an old textbook with

anyway $s=\left(\frac{1}{2}\right){ft}^{2}+Vot+Ho$
so a) $s=-16{t}^{2}+0+1362$
$\frac{ds}{dt}=velocity=-32t+0$
and of course acceleration $=\frac{{d}^{2}s}{{dt}^{2}}=-32$
b) find out how far it moved in that one second and divide by one second
$s\left(2\right)=-16\cdot 4+1362$
$s\left(1\right)=-16\cdot 1+1362$
difference $=-48$ ft in one second so average speed $=-48$ ft/second beteen
c) $v\left(1\right)=-32\left(1\right)=-32f\frac{t}{s}$
$v\left(2\right)=-32\left(2\right)=-64f\frac{t}{s}$
d) when will $s=0$?
$0=-16{t}^{2}+0+1362$
$16{t}^{2}=1362$
${t}^{2}=681$
$t=26.1$ seconds
e) $-16\left(26.1\right)=-32\left(26.1\right)$