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20 February, 09:16

Ted throws an object straight up in the air with initial velocity of 54 ft./s for a platform of 40 feet above the ground how all of the object for hits the ground

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Answers (2)
  1. 20 February, 09:41
    0
    What is the question here?
  2. 20 February, 09:43
    0
    Part 1: find how many seconds it takes for the projectile to reach it's peak (velocity = 0):

    V = V + at

    0 = 54 + - 32.2*t

    t=1.67seconds

    part 2: find how high in the air the projectile has gone (now knowing the time to peak):

    H = H + V (t) + (1/2) at^2

    H = 40+54 (1.67) + (1/2) (-32.2) (1.67^2)

    H = 85.27 feet

    part 3: find out how much time it takes to fall from 85.27 feet (peak) to the ground below

    H = H + V (t) + (1/2) at^2

    0 = 85.27 + (0) (t) + (1/2) (-32.2) (t^2)

    t = 2.3 seconds

    part4: combine the time to reach the peak plus the time to fall from peak to the ground:

    total time = 1.67 + 2.3

    total time = 4 seconds
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