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?

 To determine the time it would take for an object to fall from three times the initial height with no air resistance, we can use the kinematic equation that describes distance (d) as a function of time (t) under constant acceleration (a). The equation is:

\[ d = \frac{1}{2} a t^2 \]

If 'T' is the time it takes to fall from the initial height 'H', then we can write:

\[ H = \frac{1}{2} g T^2 \]

Here, 'g' is the acceleration due to gravity, which is approximately 9.8 m/s² near the surface of the Earth.

For three times the height, \( 3H \), the equation would be:

\[ 3H = \frac{1}{2} g t^2 \]

where 't' is the new time to fall from three times the height.

Now we want to solve for 't' when the height is '3H'. First, let's express '3H' in terms of 'H':

\[ 3 \left(\frac{1}{2} g T^2\right) = \frac{1}{2} g t^2 \]

This simplifies to:

\[ 3T^2 = t^2 \]

To solve for 't', we take the square root of both sides:

\[ \sqrt{3T^2} = t \]

\[ t = T \sqrt{3} \]

So, the time 't' it would take for the object to fall from a height three times the initial height 'H' is \( T \sqrt{3} \), assuming there is no air resistance.