A rock falls for 1.43 seconds how far did it fall? The falls's velocity is an acceleration of -9.81 m/s2

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-09-23T02:11:03+02:00

Answer:

[tex]s = -0.100 \ m[/tex]

Explanation:

Generally from kinematics equations we have that

[tex]s = ut + \frac{1}{2}gt^2[/tex]

Here u is the initial velocity of the rock and the value is 0 m/s given that there was no information that it was in motion before the fall

substituting 1.43 s for t and [tex]-9.8m/s^2[/tex] for g we have that

[tex]s = 0 *1.43 + \frac{1}{2}(-9.8)*(0.143)^2[/tex]

=> [tex]s = -0.100 \ m[/tex]

The negative sign show that the displacement is in the direction of the negative y-axis

Ritmeks Ritmeks · Answer 2 · 2021-10-15T11:52:56+02:00

The answer is -10.03 meters

The first is to write out the parameters given in the question

time= 1.43 seconds

distance(s)= ?

acceleration(g)= - 9.81

The formula is given as follows

s= ut + 1/2gt²

= (0)(1.43) + 1/2(-9.81)(1.43²)

= 0 + (-4.905)(2.0449)

= - 10.03

Hence the distance is -10.03 meters

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