Answered

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ball is thrown vertically upward at an initial speed of Its height​ (in feet) after t seconds is given by ​h(t)​t(​16t). After how many seconds does the ball reach its maximum​ height? Round answer to two decimal places.

Sagot :

If the height of ball is shown by h(t)=52t-16[tex]t^{2}[/tex] then the ball will reach after 1.625 seconds.

Given that height in t seconds is shown by h(t)=52t-16[tex]t^{2}[/tex].

We are required to find the time after which the ball will attain its maximum height.

The maximum height is the y coordinate of vertex of the parabola. Then we can use the following value of t.

h(t)=52t-16[tex]t^{2}[/tex]

Differentiate with respect to t.

dh/dt=52-32t

Again differentiate with respect to t.

[tex]d^{2}h/dt^{2}[/tex]=-32t

Because tim cannot be negative means the height is maximum.

Put dh/dt=0

52-32t=0

-32t=-52

t=52/32

t=1.625

Hence if the height of ball is shown by h(t)=52t-16[tex]t^{2}[/tex] then the ball will reach after 1.625 seconds.

Question is incomplete as the right and complete equation is

h(t)=52t-16[tex]t^{2}[/tex] .

Learn more about differentiation at https://brainly.com/question/954654

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