Uniformly accelerated linear motion downwards

[Andrei0408]

Homework Statement

A small body is freely released from a platform located 605 meters above the ground. If g=10m/s2
and
we ignore air resistance find:
a) the time required to reach the ground;
b) the velocity at the ground;
c) the
velocity at 425 m from the ground;
d) the distance traveled in the first 4 seconds of flight;
e) the distance
traveled in the last 4 seconds of flight;
f) the distance traveled in the third second of flight and in the
seventh.

Relevant Equations

motion law, velocity law

Could you check if what I did is right and help me at e) and f)? Thank you!

 

[Andrei0408]

Homework Statement:: A small body is freely released from a platform located 605 meters above the ground. If g=10m/s2
and
we ignore air resistance find:
a) the time required to reach the ground;
b) the velocity at the ground;
c) the
velocity at 425 m from the ground;
d) the distance traveled in the first 4 seconds of flight;
e) the distance
traveled in the last 4 seconds of flight;
f) the distance traveled in the third second of flight and in the
seventh.
Relevant Equations:: motion law, velocity law

Could you check if what I did is right and help me at e) and f)? Thank you!

EDIT: I tried e) again and got this:
tAD = tAO - 4
tAD = 7s
yDO = yA - 1/2 * g * (tAD)^2
yDO = 605 -245
yDO = 360m
is this right?

 

[Frigus]

•It seems to me that d. part is not correct.

•distance traveled in nth second means distance traveled between time interval of $n-1$ to $n$ seconds.

•e part—>
If total time interval is t seconds and you want to find the distance traveled in last n seconds you can calculate the the distance traveled in time interval of $t-n$ seconds and then subtract it from distance traveled in total time interval.

 

[Andrei0408]

•It seems to me that d. part is not correct.

•distance traveled in nth second means distance traveled between time interval of $n-1$ to $n$ seconds.

•e part—>
If total time interval is t seconds and you want to find the distance traveled in last n seconds you can calculate the the distance traveled in time interval of $t-n$ seconds and then subtract it from distance traveled in total time interval.

Thank you!

 

[Lnewqban]

EDIT: I tried e) again and got this:
tAD = tAO - 4
tAD = 7s
yDO = yA - 1/2 * g * (tAD)^2
yDO = 605 -245
yDO = 360m
is this right?

That result seems to be correct.

After following your work for a), b), c) and d), I believe that you take many steps that are not necessary and could introduce errors.

For resolving f), you could calculate the velocities at times 2 seconds and 7 seconds first.
Then, include each of those values into the equation to calculate the distance traveled during the following second.

 

[Andrei0408]

That result seems to be correct.

After following your work for a), b), c) and d), I believe that you take many steps that are not necessary and could introduce errors.

For resolving f), you could calculate the velocities at times 2 seconds and 7 seconds first.
Then, include each of those values into the equation to calculate the distance traveled during the following second.

Thank you, I'll try to be more careful