Question

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?

Answer 1

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 $s(t)=-16t^2+160t$

Take derivative of the distance

We know that first derivative of distance is velocity

So $v(t)=-32t+160$

At maximum height we know that velocity is zero

So $-32t+160=0$

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 = $16\times 5^2+160\times 5=1200feet$