A ball is thrown straight up from the top of a 64 foot tall building with an initial speed of 48 feet per second. The height of the ball as a function of time can be modeled by the function h(t) = –16t2 + 48t + 64. What is the maximum height of the ball ?

welp the ball is at a maximum height of 64 feet

Step-by-step explanation:

see below

Step-by-step explanation: h(t)=-16t^2+48t+64

Find when h(t) = 64

64 = -16t^2+48t+64

Subtract 64 from each side

0 =  -16t^2+48t

Factor

0 = -16t( t-3)

Using the zero product property

-16t =0    t-3 =0

t=0 and t=3

Other than at t=0,  so at t=3 seconds

How long will it take for f ball to reach its maximum height?

Find the vertex

t = -b/2a = -48 / ( 2 * -16) = -48/ -32 = 1.5 seconds

It will reach the maximum height at 1.5 seconds

What is the maximum height reached by the ball?

The height will be

h(1.5) = -16*(1.5)^2+48(1.5)+64

      = -36+72+64

      = 100

The maximum height is 100 ft

the answer is c just got it on ed

step-by-step explanation:

2/3

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