The path of a rocket is given by the equation s(t)=64t - 16t^2, where t is time and s is height.? What is the maximum height the rocket will reach? In how much time will the rocket hit the ground?

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S(t) = 64t - 16t^2

To calculate the maximum point , we need to determine the first derivative's zeros

S'(t) =64 - 32t  = 0

==> 32(2- t) = 0

==> t = 2

Then the rockect reach maximum heigth when t = 2

S(2) = 64*2 - 16(2^2)

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S(t) = 64t - 16t^2

To calculate the maximum point , we need to determine the first derivative's zeros

S'(t) =64 - 32t  = 0

==> 32(2- t) = 0

==> t = 2

Then the rockect reach maximum heigth when t = 2

S(2) = 64*2 - 16(2^2)

         = 128 - 64 = 64

Then the maximum height is S = 64

Now the time needed for the rocket to touch the ground is when the height  S = 0

==> S = 64t - 16t^2 = 0

==> 16t(4- t) = 0

==> t1 = 0

==> t2= 4

t1 = 0 is the time when the rockect starts from the ground.

t2= 4 is the time when the rocket hit the ground.

 

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