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A projectile is thrown upward so that its distance above the ground after t secondes is...
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We have given
at maximum height , its velocity will be zero.So differentiate h with respect to t
t=17.63 Sec give maximum height.
Posted by pramodpandey on May 14, 2013 at 11:05 AM (Answer #1)
We have given
at maximum height , its velocity will be zero.So differentiate h with respect to t ,so we have
`h'(t)=0 if `
t=18 Sec , give maximum height.
Posted by pramodpandey on May 14, 2013 at 11:09 AM (Answer #2)
High School Teacher
If you have quadratic function (parabola) `f(x)=ax^2+bx+c` where `a<0` (as in your case) its maximum value will be `-(b^2-4ac)/(4a)` (that is the height of vertex of parabola), and that maximum value will be reached at point `-b/(2a)` (those are seconds in your case).
So in your case `a=-12,` `b=432` and `c=0.`
Hence projectile will reach its maximum height after `-432/(2cdot(-12))=18` seconds.
That maximum value will be `h=-12cdot18^2+432cdot18=3888`
Red is maximum height, green is time in seconds till maximum height and blue is trajectory of projectile.
Posted by tiburtius on May 14, 2013 at 11:19 AM (Answer #3)
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