Question 777701
There is no initial velocity since the object,
So the equation reduces to:
{{{ h(t) = -16t^2 + h }}}
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I need to find the time interval when
{{{ h }}} goes from {{{ h = 100 }}} to {{{ h = 0 }}}
and then when
{{{ h }}} goes from {{{ h = 200 }}} to {{{ h = 0 }}}
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{{{ h = 100 }}} at {{{ t = 0 }}}
{{{ h[0] = -16*0^2 + 100 }}}
and
{{{ h[t] = 0 }}} ( when object hits ground )
{{{ 0= -16t^2 + 100}}}
{{{ 16t^2 = 100  }}}
{{{ t^2 = 100/16 }}}
{{{ t = 10/4 }}} sec
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{{{ h = 200 }}} at {{{ t = 0 }}}
{{{ h[0] = -16*0^2 + 100
and
{{{ h[t] = 0 }}} ( when object hits ground )
{{{ 0= -16t^2 + 200}}}
{{{ 16t^2 = 200  }}}
{{{ t^2 = 200/16 }}}
{{{ t = (10/4)*sqrt(2) }}} sec
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The ratio of the times is:
{{{ ((10/4)*sqrt(2)) / ( 10/4 ) = sqrt(2) }}}
The object does not take twice as long
to reach the ground. It take 1.414 times as long