Velociety 1 second later?

Question

An object with an intial velocity of 2 m/s upward is in free fall. What is it's velocity 1 second later?

Please answer and explain how you solved.
Thanks

Answer 1

v0=2m/s and t-t0=1sec
g=(v-v0)/(t-t0) for gravity
-9.81=(v-2), note g and v0 are in different
direction, so I assumed v is positive
and g is negative.
v=-7.81 m/s
which means it is going downwards

Answer 2

accelaration due to gravity is 10m/s^2 towards the earth

initial velocity is 2m/s upwards (or if you think in terms of velocity towards the earth it is -2m/s as its moving away)

in one second it would have accelerated by 10m/s towards the earth.

Adding this two velocities (remembering they are in different directions) gives

final speed (downwards) = -2 + 10 = 8

speed one second later is 8 m/s towards the ground

hope that helps

Answer 3

use the equation for constant acceleration:
v(final) = v(initial) + at
= -2m/s + (9.8m/s^2)(1sec)
=7.8 m/s
be carefully about the signs of the values. +ve is for down while -ve is up i.e free fall (g) is directed down while the initial velocity is up.The result velocity is +ve which means that it is directed downwards so its on its way down affter that time

Answer 4

Is this a trick ?
an object in free fall there is no up or down .It should stay at
2 m/s or I could be wrong and you are smarter than me.

Answer 5

a= dv/dt = -g
v = -gt +c
When t = 0, v = +2 m/s
So v= -gt +2 = -9.8t +2
1 second later v = -9/8 +2 = -7.8 m/sec