An object is thrown downward from the top of a building. After 𝑡 seconds, the object’s height (in feet above the ground) is given by the formula 𝑠(𝑡)=−16𝑡2

Question

An object is thrown downward from the top of a building. After 𝑡 seconds, the object’s height (in feet above the ground) is given by the formula 𝑠(𝑡)=−16𝑡2−30𝑡+190. a) Determine the object’s average velocity on the interval [𝑎,𝑎+ℎ]. Simplify your answer completely. b) Determine the object’s instantaneous velocity after one second and after two seconds. c) Determine when the object hits the ground. d) Determine the object’s instantaneous speed as it hits the ground.

Answer 1

 

 

a)   v(t) = s'(t) = -32t - 30

 

        v(a+h) - v(a)         -32(a+h) - 30   - (-32a - 30)

       ----------------- = --------------------------------------  

            (a+h)-a                    a + h - a

 

                           = -32a - 32h - 30 + 32a + 30

                              ---------------------------------

                                               h

 

                            =    -32h/h

 

                            = -32

 

 

B) instantaneous velocities

V(1) = -32(1) - 30 = -32 - 30 = -62

 

V(2) = -32(2)  - 30 = -64 - 30 = -94

 

 

C)  -16t^2 - 30t + 190 = 0

 

      8t^2 + 15t - 95 = 0

 

     Quadratic formula says:

         t = -15 +or- sqrt( 15^2 - 4(8)(-95)) / 16

 

     the negative case results in negative times.

 

    t = (-15 + 57 and 1/7)/16 = 2.634

 

v(2.634) = -32*2.634 - 30 = -114.288