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If a 167 kg object falls from a height of 550 meters, what is the acceleration of that object? (Assume that the initial velocity is 1 m/s and we use G (9.8m/s^2))

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Luv2Teach

Answer:

Explanation:

I used 2 different equations to solve this. First,

Δx = v₀t + [tex]\frac{1}{2}[/tex]at² which says that the displacement of an object is equal to its initial velocity times the time in seconds plus one-half the acceleration of the object times the time squared. I solved this for t then plugged into

[tex]a=\frac{v_f-v_0}{t}[/tex] Let me first say that I used 1.0 as the velocity in the first equation since 1 is too insignificant. Filling into the first equation and using the fact that the object lands 550 meters below the point at which it starts (making the displacement negative):

[tex]-550=1.0t+(-9.8)t^2[/tex] which simplifies a bit to

[tex]-550=1.0t-4.9t^2[/tex] and then get everything on one side and factor:

[tex]-4.9t^2+1.0t+550=0[/tex] gives us that

t = 10.7 seconds which should be rounded to 11 seconds. I'm going to use t = 11 here.

[tex]a=\frac{0-1.0}{11}[/tex] so

a = -.091 m/s/s

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