Standing on top of a building, Bob tosses a baseball straight up with an initial speed of 15 m/s. It takes 4.0 s for the ball to hit the gro

Question

Standing on top of a building, Bob tosses a baseball straight up with an initial speed of 15 m/s. It takes 4.0 s for the ball to hit the ground. What is the vertical distance of the building? (g = -9.8 m/s²)

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Thiên Ân 4 years 2021-08-27T19:26:00+00:00 1 Answers 26 views 0

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  1. thienthi
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    2021-08-27T19:27:16+00:00
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    Answer:

    48.26 m

    Explanation:

    time to goes up (till stop for a while in the air – maximum height)

    vt = vo + a t

    0 = 15 + g . t

    0 = 15 + (-9.8) . t

    9.8t = 15

    t = 1.531 s

    so the time left to goes down is

    4.0 – 1.531 = 2.469 s

    height from the top of building can find it by using

    vo =√(2gh)

    15 = √(2)(9.8).h

    15² = 19.6h

    h = 225/19.6 = 11.48 m

    so the distance of maximum height to the ground is

    t = √(2H/g)

    2.469 = √(2H/9.8)

    2.469² = 2H/9.8

    6.096 = 2H/9.8

    2H = 6.096 x 9.8 = 59.74 m

    so the vertical distance of the building (or the building height’s is)

    H – h = 59.74 – 11.48 = 48.26 m

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