Finding height in uniformly accelerated motion

[greg_rack]

Homework Statement

An object is thrown vertically upwards from ground level with an initial velocity of 40 ms–1.
2.0 seconds later another object is released from a height above the ground and falls vertically
from rest.
Both of the objects hit the ground at the same time.
From what height above the ground was the second object released?

Relevant Equations

$v=v_{0}+at$
$\Delta x=\frac{v^2-v_{0}}{2a}$

First of all I have found the time taken by the first object to hit the ground back: $\Delta t=2(\frac{v}{a})$.
Then, by subtracting 2 seconds to this quantity, I get the time in which the second object has accelerated in free fall, with terminal velocity $v=at$.
Now, I find the distance traveled by the second object: $\Delta x=\frac{v}{2a}$... but the result I get is wrong!

 

[Lnewqban]

What value of height your result has?
Could you show us your work?

 

[LCSphysicist]

Homework Statement:: An object is thrown vertically upwards from ground level with an initial velocity of 40 ms–1.
2.0 seconds later another object is released from a height above the ground and falls vertically
from rest.
Both of the objects hit the ground at the same time.
From what height above the ground was the second object released?
Relevant Equations:: $v=v_{0}+at$
$\Delta x=\frac{v^2-v_{0}}{2a}$

First of all I have found the time taken by the first object to hit the ground back: $\Delta t=2(\frac{v}{a})$.
Then, by subtracting 2 seconds to this quantity, I get the time in which the second object has accelerated in free fall, with terminal velocity $v=at$.
Now, I find the distance traveled by the second object: $\Delta x=\frac{v}{2a}$... but the result I get is wrong!

"First of all I have found the time taken by the first object to hit the ground back: $\Delta t=2(\frac{v}{a})$".Ok
"Then, by subtracting 2 seconds to this quantity, I get the time in which the second object has accelerated in free fall, with terminal velocity $v=at$. ?" Terminal velocity? Where is the air resistance to you talk about it?
"Now, I find the distance traveled by the second object: $\Delta x=\frac{v}{2a}$... but the result I get is wrong!" This equation is wrong, you forget one square

Why don't you try to use the simple $h = 0.5at^2$? You don't need to know more than the time

 

[greg_rack]

"First of all I have found the time taken by the first object to hit the ground back: $\Delta t=2(\frac{v}{a})$".Ok
"Then, by subtracting 2 seconds to this quantity, I get the time in which the second object has accelerated in free fall, with terminal velocity $v=at$. ?" Terminal velocity? Where is the air resistance to you talk about it?
"Now, I find the distance traveled by the second object: $\Delta x=\frac{v}{2a}$... but the result I get is wrong!" This equation is wrong, you forget one square

And that's exactly the square which was missing... Now it works! Thank youu