Managerial Calculus (An object shot into the air from a position )

An object shot into the air from a position on the edge of a cliff next to the sea moves so that its height above sea level at time t seconds is given by the formula s(t) = -16t^2 + 80t +96 where s is measured in feet. Find:

(a) when does it reach its highest point?
(b) How high does it go?
(c) What is the average velocity for t between t = 1 and t = 4?
(d) how fast is it going at t = 3?

Q. An object shot into the air from a position on the edge of a cliff next to the sea moves so that its height above sea level at time t seconds is given by the formula s(t) = -16t^2 + 80t +96 where s is measured in feet. Find:

(a) when does it reach its highest point?
Sol differentiate
velocity= ds/dt = -32t +80
At the highest place velocity is zero
Therefore -32t +80 =0
t = 80/32 = 5/2 = 2.5 seconds.............Ans

(b) How high does it go?
Sol: put t = 5/2 in s(t) = -16t^2 + 80t +96
s(2.5) = -16(2.5)^2 +80(2.5) +96
s = -100 +200+96= 196 feet.......................Ans

(c) What is the average velocity for t between t = 1 and t = 4?
s(1) = -16(1)^2+80(1)+96 = -16+80+96= 160......(i)
s(4) = -16(4)^2+80(4)+96= -256+320+96 = 160....(ii)
Average velocity = (160+160)/2 = 160 feet/sec.................Ans
(d) how fast is it going at t = 3?
velocity = ds/dt = -32t +80
put t = 3
velocity = -32*(3) +80 = -96+80 = -16 feet/sec
It is going up at 16 feet/sec................Ans

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