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If an object is projected upward from ground level with an initial velocity of 160 ft per​ sec, then its height in feet after t seconds is given by
s(t)=-16t^2+ 160t. Find the number of seconds it will take to reach its maximum height. What is this maximum​ height?

Respuesta :

Answer:

Time will be 5 sec

And maximum height will be 1200 feet    

Explanation:

We have given that initial velocity of the object u = 160 ft/sec'

Function of height [tex]s(t)=-16t^2+160t[/tex]

Take derivative of the distance

We know that first derivative of distance is velocity

So [tex]v(t)=-32t+160[/tex]

At maximum height we know that velocity is zero

So [tex]-32t+160=0[/tex]

t = 5 sec

So it will take 5 sec to reach maximum height

Now maximum height will occur at t = 5 sec

So maximum height = [tex]16\times 5^2+160\times 5=1200feet[/tex]

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