An object is thrown straight up with a velocity, in ft/s, given by v(t)= -32t + 83, where t is in seconds, from a height of 46 feet.
a) what is the object's initial velocity?
b) what is the object's maximum velocity?
c) what is the object's maximum displacement?
d) when does the maximum displacement occur?
e) when is the object's displacement 0?
f) what is the object's maximum height?

Explanation:

We have velocity

             v(t)= -32t + 83

Integrating

              s(t) = -16t²+83t+C

At t = 0 displacement is 46 feet

              46 = -16 x 0²+83 x 0+C

                 C = 46 feet

So displacement is

              s(t) = -16t²+83t+46

a) Initial velocity is

                 v(0)= -32 x 0 + 83 = 83 ft/s

       Initial velocity = 83 ft/s

b) Maximum velocity is when the object reaches ground, that is s(t) = 0 ft

Substituting

             0 = -16t²+83t+46

             t = 5.70 seconds

Substituting in velocity equation

           v(t)= -32 x 5.70 + 83 = -99.4 ft/s

           Object's maximum speed = 99.4 ft/s

c) Maximum displacement is when the velocity is zero

   That is

                 -32t + 83 = 0

                       t = 2.59 s

Substituting in displacement equation

                s(2.59) = -16 x 2.59²+83 x 2.59+46 = 153.64 ft

Object's maximum displacement = 153.64 ft

d) Maximum displacement occur at t = 2.59 seconds.

e) Refer part b

   The displacement is zero when t = 5.70 seconds

f) Same as option d

   Object's maximum height = 153.64 ft

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

false

explanation: