Answered

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A projectile is fired into the air with an initial vertical velocity of 160 ft/sec from ground level. How many seconds later does the projectile reach the maximum height? (numerical answer only)

h(t)=−16t2+160t

Sagot :

MrRoyalidn

The maximum height of the projectile is the maximum point that can be gotten from the projectile equation

The projectile reaches the maximum height after 5 seconds

The function is given as:

[tex]\mathbf{h(t) = -16t^2 + 160t}[/tex]

Differentiate the function with respect to t

[tex]\mathbf{h'(t) = -32t + 160}[/tex]

Set to 0

[tex]\mathbf{h'(t) = -32t + 160 = 0}[/tex]

So, we have:

[tex]\mathbf{-32t + 160 = 0}[/tex]

Collect like terms

[tex]\mathbf{-32t =- 160 + 0}[/tex]

[tex]\mathbf{-32t =- 160}[/tex]

Solve for t

[tex]\mathbf{t = \frac{- 160}{-32}}[/tex]

[tex]\mathbf{t = 5}[/tex]

Hence, the projectile reaches the maximum after 5 seconds

Read more about maximum values at:

https://brainly.com/question/6636648

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