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?
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.
s = 64 t - 16 t^2 = 0
==> 16t(4 - t) = 0
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
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.
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.