What times it takes for the ball to drop?

[Serious Max]

Homework Statement

When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it from three times that height, how long (in terms of T) would it take to reach the ground?

Homework Equations

$x=x_0+v_{0x}t+\frac{1}{2}a_x t^2$

The Attempt at a Solution

Did I solve it correctly?

$0=h+\frac{1}{2}(-g)T^2$

$h=\frac{1}{2}gT^2$

Now we take 3 times the height and substitute it:

$0=3\cdot\frac{1}{2}gT^2+\frac{1}{2}(-g)t^2$

$t^2=\dfrac{3}{2}gT^2\cdot \dfrac{2}{g}$

$t^2=3T^2$

$t=\sqrt{3T^2}$

 

[ehild]

It is correct. Do you have any doubts?

 

[SteamKing]

Homework Statement

When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it from three times that height, how long (in terms of T) would it take to reach the ground?

Homework Equations

$x=x_0+v_{0x}t+\frac{1}{2}a_x t^2$

The Attempt at a Solution

Did I solve it correctly?

$0=h+\frac{1}{2}(-g)T^2$

$h=\frac{1}{2}gT^2$

Now we take 3 times the height and substitute it:

$0=3\cdot\frac{1}{2}gT^2+\frac{1}{2}(-g)t^2$

$t^2=\dfrac{3}{2}gT^2\cdot \dfrac{2}{g}$

$t^2=3T^2$

$t=\sqrt{3T^2}$

Or, simplifying, $t=\sqrt{3} T$

 

[Serious Max]

Just checking. I solved it incorrectly a couple times. And actually corrected it as I was posting it here, when I realized I made a mistake once again.

SteamKing, right. That's better :).Thanks.