Question 762491
{{{ h(t) = -16t^2 - 32t + 400 }}}
First of all, I'll show you the plot of this equation:
{{{ graph( 400, 400, -2, 6, -50, 450, -16x^2 - 32x + 400 ) }}}
You can see the height of the building at {{{ t = 0 }}}
{{{ h(0) = -16(0)^2 - 32*0 + 400 }}}
{{{ h(0) = 400 }}}
So, the height of the ball is the same as the height of the building
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The problem doesn't ask you for it, but you can also see that
the ball hits the ground at about {{{ t = 4.1 }}} sec or so.
The height is zero, so {{{ h(t) = 0 }}}
Solving:
{{{ h(t) = -16t^2 - 32t + 400 }}}
{{{ 0 = -16t^2 - 32t + 400 }}}  
{{{ -t^2 - 2t + 25 = 0 }}}
Using quadratic formula:
{{{ t = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = -1 }}}
{{{ b = -2 }}}
{{{ c = 25 }}}
{{{ t = (-(-2) +- sqrt( (-2)^2 - 4*(-1)*25 )) / (2*(-1)) }}}
{{{ t = ( 2 +- sqrt( 4 + 100 )) / (-2) }}}
{{{ t= ( 2 - sqrt( 104 ) ) / ( -2) }}}
{{{ t= ( 2 - 10.198 ) / ( -2) }}}
{{{ t = ( -8.198 ) / (-2) }}}
{{{ t = 4.099 }}}
My guess was pretty good
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They are asking you for {{{ t }}} when {{{ h(t) = 96 }}}
{{{ h(t) = -16t^2 - 32t + 400 }}}
{{{ 96 = -16t^2 - 32t + 400 }}}
{{{ -16t^2 - 32t + 304 = 0 }}}
{{{ -4t^2 - 8t + 76 = 0 }}}
{{{ -t^2 - 2t + 19 = 0 }}}
Use quadratic formula
{{{ t = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = -1 }}}
{{{ b = -2 }}}
{{{ c = 19 }}}
{{{ t = (-(-2) +- sqrt( (-2)^2 - 4*(-1)*19 )) / (2*(-1)) }}}
{{{ t = ( 2 +- sqrt( 4 + 76 )) / (-2) }}}
{{{ t = ( 2 +- sqrt( 80 )) / (-2) }}}
{{{ t = ( 2 +- sqrt( 16*5 )) / (-2) }}}
{{{ t = ( 2 - 4*sqrt( 5 ) ) / (-2) }}}
{{{ t = 2*sqrt(5) - 1 }}}
{{{ t = 3.472 }}} sec
This seems to agree with the plot
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Just for explanation, suppose the {{{ -32t }}} had been +
{{{ h(t) = -16t^2 + 32t + 400 }}}
Here's the plot:
{{{ graph( 400, 400, -2, 7, -50, 500, -16x^2 + 32x + 400 ) }}}
This is the case where the ball is thrown upward instead of down.
Hope all this helps