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17 July, 21:16

A ball is thrown straight up from the top of a 24 foot tall building with an initial velocity of

40 feet per second. The height of the ball as a function of time can be modeled by the function

h = - 16t2 + 40t + 24

Part A

What is the height of the ball after 1 second? h=

Part B

How long will it take for the ball to hit the ground (set h=0 and factor completely)

seconds

+2
Answers (1)
  1. 17 July, 21:21
    0
    Step-by-step explanation:

    Part A

    The height of the ball as a function of time can be modeled by the function

    h = - 16t² + 40t + 24

    When t = 1 second,

    h = - 16 * 1² + 40 * 1 + 24

    h = - 16 + 40 + 24

    h = 48 feet

    Part B

    The equation would be

    h = - 16t² + 40t + 24 = 0

    Dividing both sides of the equation by 4, it becomes

    - 4t² + 10t + 6 = 0

    We would find two numbers such that their sum or difference is 10t and their product is - 24t².

    The two numbers are 12t and - 2t. Therefore,

    - 4t² + 12t - 2t + 6 = 0

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

    t - 3 = 0 or - 4t - 2 = 0

    t = 3 or t = 2 / - 4 = - 1/2

    Since the time cannot be negative, then t = 3 seconds

    It will take 3 seconds for the ball to hit the ground.
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