# 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)

= 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**.

The height s is given by the function:

f(s) = 64t - 16t^2

Where: t = time

The height will be maximum at a point where:

f'(s) = 0

We find the corresponding value of t value as follows.

f'(s) = 64 - 2*16t = 64 - 32t

Equating f('s) to 0

64 - 32t = 0

==> -32t = - 64

==> t = -64/-32 = 2

we get the maximum height by substituting this value of t in equation of height:

Maximum height = f(2) = 64*2 - 16*(2^2) = 128 - 64 = 64

When the rocket hits the ground s = 0

Therefor to find the time when the rocket hits the ground we equate the equation for height to 0.

Thus:

s = 64 t - 16 t^2 = 0

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

Therefore:

t = 0 and t = 4

t = 0 represents the time when the rocket is fired from ground level, and t = 4 represents the time when the rocket again falls to the ground level

Answer:

Maximum height of rocket = 64 units

Time taken to hit the ground = 4 units

The path function is s(t) = 64t-16t^2, t is time and s (t) is the height at any time t.

Solution:

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

s(t) = - ( -2*8*4t+16t^2)

s(t) = 8^2- (8^2-2*8*4t+4t^2)

s(t) = 8^2-(8-4t)^2 which which could be greatest = 64 when 8=4t, as (8-4t)^2 being a square is always positive and least for 8-4t = 0. Or t = 8/4 = 2seconds.

So greatest heiht of the rocket is s(2) = 64*2-14*2^2 = 128-64= 64 at time t = 2.