# What times it takes for the ball to drop?

1. Mar 6, 2015

### maxpan

1. The problem statement, all variables and given/known data
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?

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

3. 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:

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

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

$t^2=3T^2$

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

2. Mar 6, 2015

### ehild

It is correct. Do you have any doubts?

3. Mar 6, 2015

### SteamKing

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

4. Mar 6, 2015

### maxpan

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.