Jason jumped off a cliff into the ocean in Acapulco while vacationing with some friends. His height as a function of time could be modeled by the function h ( t ) = − 16 t 2 + 16 t + 480 h(t)=−16t 2 +16t+480, where t is the time in seconds and h is the height in feet. What is the maximum height that Jason reaches?

Answer:

484feet

Step-by-step explanation:

If the height as a function of time could be modeled by the function h(t)=−16t^2 +16t+480, where t is the time in seconds and h is the height in feet.

The maximum height of Jason occurs when dh/dt = 0 (velocity is zero)

dh/dt = -32t + 16

0 = -32t + 16

32t = 16

t = 16/32

t = 1/2

t = 0.5secs

Get the maximum height

Recall that h(t)=−16t^2 +16t+480

Substitute t = 0.5

h(0.5)=−16(0.5)^2 +16(0.5)+480

h(0.5) = -4 + 8 + 480

h(0.5) = 4 + 480

h(0.5) = 484

Hence the maximum height reached by Jason is 484feet

The maximum height that Jason reaches is h =  484 feet and it will be reached at t = 0.5 sec.

Let the function be denoted by [tex]y = f(x)[/tex]

If it is twice differentiable, then, firstly, we differentiate it with respect to x and equate with 0 to find the critical values.

Let the obtained critical values be [tex]x_1, x_2, ... x_n[/tex]

The second derivative of that function is then evaluated on those critical values.

The [tex]i^{th}[/tex] critical value has got the maximum if [tex]f''(x_i) < 0[/tex]

That means, if at  [tex]x = x_i[/tex] , we get

[tex]f'(x-i) = 0\\and\\f''(x_i) < 0[/tex]

Then that point is where function will have maximum.
If [tex]f''(x_i) > 0[/tex], then the point where the function will have minimum

If value of second rate at point [tex]x_i[/tex] is 0, then we go for third rate of function and check the same facts so on for upper rate(if they exist).

For the given case, we're given the height function as:

[tex]h(t) = - 16t^2 + 16t + 480[/tex]

The function is infinitely differentiable as its polynomial(by a theorem).

Its first and second rate with respect to 't', we get;

[tex]h(t) = - 16t^2 + 16t + 480\\h'(t) = -32t + 16\\h''(t) = -32 < 0[/tex]

Thus, all critical points will be maximum points.

The critical points are evaluated by [tex]h'(t) = 0[/tex]

They are calculated as:

[tex]h'(t) = 0\\-32t + 16 = 0\\t = 0.5 \: \rm sec[/tex]

The height at t = 0.5 s is evaluated as:

[tex]h(0.5) = - 16(0.5)^2 + 16(0.5) + 480 = 484 \: \rm feet[/tex]

Thus, at time 0.5, the height function will be at its maximum value(484 feet). That means, the height of Jason will be maximum when time will be 0.5 seconds from initial time.

Learn more about maximum and minimum values here:

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