Question 562522
{{{ x(t) = -16t^2 + 400 }}}
Note that:
{{{ x(0) = -16*0 + 400 }}}
{{{ x(0) = 400 }}} ft
This just says that at {{{ t = 0 }}},
the ball is {{{ 400 }}} ft above ground
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I need to find when the ball is at the
ground, or when {{{ x = 0 }}}
{{{ 0 = -16t^2 + 400 }}}
{{{ 16t^2 = 400 }}}
{{{ t^2 = 400/16 }}}
{{{ t^2 = 25 }}}
{{{ t = sqrt(25) }}}
{{{ t = 5 }}} sec