Question 1087912
For the question:
" After how many seconds will the projectile be 100 m above the ground? "
the units are meters and seconds, so all you need is the formula
{{{ d = -4.9t^2 + 73.5t }}}
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The formula is a parabola with a peak, so unless the projectile is at its
peak at 100 m above ground, there will be 2 solutions, one when the projectile is
on the way up, and another on the way down.
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{{{ d = 100 }}} m
Plug this value into the formula
{{{ 100 = -4.9t^2 + 73.5t }}} 
{{{ -4.9t^2 + 73.5t - 100 = 0 }}} 
Use the quadratic formula
{{{ t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = -4.9 }}}
{{{ b = 73.5 }}}
{{{ c = -100 }}}
{{{ t = (-73.5 +- sqrt( 73.5^2-4*(-4.9)*(-100) ))/(2*(-4.9)) }}}
{{{ t = (-73.5 +- sqrt( 5402.25 - 1960 ))/(-9.8) }}}
{{{ t = (-73.5 +- sqrt( 3442.25 ))/(-9.8) }}}
{{{ t = ( -73.5 + 58.671 ) / (-9.8 ) }}}
{{{ t = 14.829  / 9.8 }}}
{{{ t = 1.513 }}} sec
and, the other solution is:
{{{ t = ( -73.5 - 58.671 ) / (-9.8 ) }}}
{{{ t = 132.171 / 9.8 }}}
{{{ t = 13.487 }}} sec
These are the 2 solutions
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" How long will it take for the projectile to return to the ground? "
At ground level, {{{ d = 0 }}}
{{{  -4.9t^2 + 73.5t = 0 }}} 
{{{ t*( -4.9t + 73.5 ) = 0 }}}
{{{ t = 0 }}} This is when the projectile is launched
{{{ -4.9t + 73.5 = 0 }}}
{{{ 4.9t = 73.5 }}}
{{{ t = 15 }}} 
In 15 sec, the projectile hits the ground.
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I can use this answer to check my 1st answer.
The peak will be at 1/2 this time, or {{{ 7.5 }}} sec
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My solutions for the height of {{{ 100 }}} m must be
equally spaced on either side of {{{ 7.5 }}} sec, so I can say:
{{{ 7.5 - 1.513 = 13.487 - 7.5 }}}
{{{ 5.987 = 5.987 }}}
Right on the money!
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Here is a plot of this problem:
{{{ graph( 400, 400, -2, 16, -20, 300, -4.9x^2 + 73.5x ) }}}