You throw an object up with a speed of v0y = 9 m/s from a height of y = 25 m. a) How long, in seconds, does it take for the obj

Question

You throw an object up with a speed of v0y = 9 m/s from a height of y = 25 m.

a) How long, in seconds, does it take for the object to reach the ground?

b) What is the object’s final velocity, in meters per second, as it impacts the ground?

c) Find in the time, in seconds, if you instead threw the object down the same velocity, Voy

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Cherry 3 years 2021-07-27T09:15:03+00:00 1 Answers 46 views 0

Answers ( )

  1. RuslanHeatt
    0
    2021-07-27T09:17:00+00:00
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    Answer:

    a) 3.36 sec

    b) 23.9 m/s

    c) 1.52 sec

    Explanation:

    maximum height

    H = 25 +v_y^2/2g

    H=25+9^2/2*9.81 =29.13 m

    time taken to reach maximum height = t_1 =v_y/g = 9/9.81 =0.92 seconds

    time taken to fall back to ground from maximum height = √(2gH)/g

    = √(2*9.81*29.13)/9.8 = 2.439 seconds

    a) Total time taken to reach ground T = t_1+t_2 =0.92+2.439=3.36 sec

    b) objects final speed = √2gH = 23.9 m/s

    c) let total time be t then

    25 = 9t +0.5gt^2

    Solving we get t= 1.52 seconds

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