Question

all frictional forces are considered to be negligible. what distance does the object fall 10 s after it is released from rest?

Answer 1

t=10s (provided in the question)

u=0m/s

the initial velocity is taken as 0 as the body is starting from rest even if it is falling.

s=?

a=-g =-9.8 (acceleration due to gravitation)

Let v be final velocity.

Where 't' represents time, 's' represents distance, 'u' is initial velocity while 'g' is the acceleration due to gravity.

Now using the second equation of Newton's equations of Motion.

s=ut+1/2a$t^{2}$

Now putting values into the equation above:

s=(0)(10)+1/2(-9.8)x$(10)^{2}$

s=0-1/2x9.8x100

s=-4.9x100

s=-490m

That is the distance is 490m where the negative sign represents the direction of the distance towards the earth instead of away from it.

Learn more about Newtons Equations of Motion at:

brainly.com/question/18827799

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