A ball is thrown straight up from the top of a building 144 ft tall with an initial velocity of 48 ft per second. The height s(t) (in feet)
of the ball from the ground, at time t (in seconds), is given by
s(t) = 144 + 48t − 16t^2.
Find the maximum height attained by the ball.

180feet

Step-by-step explanation:

Given the function that models the height of a ball

s(t) = 144 + 48t − 16t^2.

At maximum height, the velocity is zero ie. ds/dt  = 0

ds/dt = 48 - 32t

0 = 48 - 32t

48 = 32t

t = 48/32

t = 1.5secs

Get the maximum height

s(t) = 144 + 48t − 16t^2.

s(1.5) = 144 + 48(1.5)-16(1.5)^2

s()1.5) = 144 + 72 - 16(2.25)

s(1.5) = 144 + 72-36

s(1.5) = 180 feet

Hence the maximum height attained by the ball is 180feet

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step-by-step explanation:

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