The position of an object that is projected straight up from the ground with an initial velocity of 114 ft/sec is given by s(t) = 114t − 16t2^

Question

How do I find, At what time (in seconds) does the object reach its maximum height? (Round your answer to one decimal place.)

When my teacher ask this At what time (in seconds) does the object strike the ground? does it me at what time the equation equals 0?

Answer 1

s(t) = 114t -16t^2

s'(t) = 114-32t =0

t = 114/32 = 57/16 = 3 9/16 seconds to reach maximum height

s(57/16) = 114(57/16) -16(57/16)^2

= 57(57)/8 - 57/16

=406 2/16 - 203 1/16 feet = 203 1/16 feet max height

s(t) = 114t-16t^2 = 0

t(114-16t) = 0

t = 114/16 = 57/8 = 7 1/8 seconds to hit the ground

maximum height is reached in exactly half that time = 7 1/8 divided by 2 = 3 9/16 seconds

the equation is a downward opening parabola with x or t intercepts of (0,0) and (7 1/8, 0)

the parabolas' vertex = the maximum = (3 9/16, 203 /16) = (3.5625, 203.0625)