Basic free fall It takes 0.210 seconds for a dropped object to pass a window that is 1.35 meters tall. From what height above the top of the window was the object released? Air resistance is negligible.
I get 1.5 roughly as an answer. What I do is $D = 0.5at^2+v_{\rm initial}t$
to find the initial velocity.
Then find out how long it takes to accelerate to that velocity: $V_{\rm final} = at$
Then I plug in that time in the distance formula with the acceleration of gravity $D = at^2$,
and I get 1.5. 
Is this answer correct? Is this method efficient. My teacher said that it is inefficient and that there is a faster way. What is the faster way? Thanks so much!
 A: For an "exact" solution, we can also use the following variant of the argument of Henning Makholm.  
Let the height above the window be $h$, let the total time to reach the top of the window be $t_1$, and total time to reach the bottom of the window be $t_2$.  Let acceleration due to gravity be constant at $a=9.8$.
Then 
$$h=\frac{1}{2}at_1^2 \qquad\text{and}\qquad h+1.35=\frac{1}{2}at_2^2.$$
Subtract. We get 
$$\frac{1}{2}a(t_2^2-t_1^2)=1.35.$$
But $t_2^2-t_1^2=(t_2-t_1)(t_2+t_1)$. Since $t_2-t_1=0.210$ we obtain 
$$t_2+t_1=\frac{(2)(1.35)}{(0.210)(9.8)}.$$
Note that $t_1=(1/2)((t_2+t_1)-(t_2-t_1))$. Use our expression for $t_2+t_1$,   together with $t_2-t_1=0.210$, to find $t_1$. Now we know $t_1$, so we know $h$.   
Another way: It is more pleasant to avoid numbers until the end.  Let $w$ be the height of the window, and let $s$ be the amount of time it took for the object to pass the window. Let $u$ be the velocity at the top of the window, and 
$v$ the velocity at the bottom.  Then the average velocity at which the window was traversed is $(v+u)/2$.  But it is also $w/s$, and therefore
$$\frac{v+u}{2}=\frac{w}{s}.$$
The change in velocity is $v-u$. It is also $as$. Thus
$$\frac{v-u}{2}=\frac{as}{2}.$$
From the above two equations it follows that 
$$u=\frac{w}{s}-\frac{as}{2}.$$
The average velocity from the time of dropping until arriving at the window is $u/2$.  The time it took is $u/a$, so the distance travelled is $u^2/2a$.  It follows that the height above the window from which the object was dropped is
$$\frac{\left(\frac{w}{s}-\frac{as}{2}\right)^2}{2a}.$$ 
A: It's not quite clear to me what you're doing I suspect you're assuming that the vertical speed is constant while the object passes the window, which is not accurate. Here's what I would do:
The dropped object follows a parabola:
$$h(t) = p + qt - (g/2)t^2$$
for some coefficients $p$ and $q$ to be determined.
Declare the top of the window to be at zero elevation, and the moment at which the object passes that point to be $t=0$. Then $h(0)=p=0$, giving us one of the coefficients.
Because $h(0.21)=-1.35$ we can solve for $q$ and get
$$0.21 q - (g/2)(0.21^2) = -1.35$$
$$q = \frac{(g/2)(0.21^2) - 1.35}{0.21} \approx -5.40$$
Finally find the apex of the parabola. The roots of the polynomial are $0$ and $\frac{q}{g/2}$, so the object was dropped at time $q/g$ from height $h(q/g)=\frac{q^2}{2g} \approx 1.48$.
