Question

A brick is dropped from a big scaffold. What is its velocity after 4.0s?

Answer 1

The velocity of the brick is 39.2 m/s downward

Explanation:

The motion of the brick is a free fall motion, since the object is affected only by the force of gravity. Therefore, it has a uniformly accelerated motion towards the ground, with constant acceleration of $g=9.8 m/s^2$.

So, we can find its velocity using the suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the brick in this problem (taking downward as positive direction)

u = 0 (it is dropped from rest)

$a=g=9.8 m/s^2$

Therefore, its velocity after t = 4.0 s is:

$v=0+(9.8)(4.0)=39.2 m/s$

Downward, because the sign is positive.

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