Question

un ladrillo se deja caer de un andamio alto. Cual es su velocidad despues de 4.0 s? Que tanto cae el ladrillo durante este tiempo?​

Answer 1

1) The velocity of the object after 4.0 s is 39.2 m/s

2) The distance covered by the object is 78.4 m

Explanation:

1)

The object is in free fall (the only force acting on it is the force of gravity), so it is a uniformly accelerated motion, with constant acceleration $g=9.8 m/s^2$ towards the ground. Therefore, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem, we have:

u = 0 (because the object is dropped from rest)

$a=g=9.8 m/s^2$ is the acceleration

Substituting t = 4.0 s, we find the velocity of the object after 4.0 seconds:

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

2)

The distance covered by the object during this time interval can be found by using another suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem, we have:

u = 0

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

And substituting t = 4.0 s, we find the distance covered by the object during the fall:

$s=0+\frac{1}{2}(9.8)(4.0)^2=78.4 m$

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