A ball is thrown strait 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)=-16t^2+48t+64. At what other time will the ball be at a height of 64 feet? How long will it take for f ball to reach its maximum height? What is he maximum height reached by the ball? How long willlit take for the ball to hit the ground?

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

How long will it take for the ball to hit the ground?

h(t) =0

0=-16t^2+48t+64

Factor

0 = -16( t^2 - 3t -4)

0=- 16( t-4) (t+1)

Using the zero product property

t-4=0     t+1=0

t =4     t = -1

Since time cannot be negative

t=4 seconds

what is the approximate angle of acb if abc is 25cm and the length of segment bc is 15

step-by-step explanation: