Question 1076627
{{{ h(t) = -(1/2)*9.81t^2 + 2.1t + h[0] }}}
When it hits the ground, {{{ h(t) = 0 }}}
{{{ 0 = -(1/2)*9.81t^2 + 2.1t + h[0] }}}
It takes {{{ t = 3 }}} sec to hit the ground, so
{{{ 0 = -(1/2)*9.81*3^2 + 2.1*3 + h[0] }}}
{{{ 44.145 - 6.3 = h(0) }}}
{{{ h(0) = 37.845 }}} m
The balcony is 37.845 m high
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{{{ h^2(3) = (-1/2)*9.81t + 2.1 }}}
{{{ h^2(3) = (-1/2)*9.81*3 + 2.1 }}}
{{{ h^2(3) = -14.7 + 2.1 }}}
{{{ h^2(3) = -12.6 }}}
The flowerpot was moving 12.6 m/sec in
the down direction just before it hit ground
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Here is the graph:
{{{ graph( 400, 400, -1, 5, -5, 50, (-1/2)*9.81*x^2 + 2.1x + 37.845 ) }}}
To check this I'll find {{{ h[max] }}}
{{{ t[max] = -b/(2a) }}}
{{{ t[max] = -2.1/(2*(-4.9)) }}}
{{{ t[max] = .2143 }}} sec
and
{{{ h[max] = -(1/2)*9.81*.2143^2 + 2.1*.2143 + 37.845 }}}
{{{ h[max] = -(1/2)*9.81*.0459 + .45 + 37.845 }}}
{{{ h[max] = -(1/2)*.45 + .45 + 37.845 }}}
{{{ h[max] = .225 + 37.845 }}}
{{{ h[max] = 38.07 }}} m
This looks about right
Definitely get a 2nd opinion on this!