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!
Best Answer
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 $$\frac{\left(\frac{w}{s}-\frac{as}{2}\right)^2}{2a}.$$