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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!

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2 Answers

up vote 4 down vote accepted

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

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"It is more pleasant to avoid numbers until the end." - because you then obtain something you can use when you encounter similar problems the next time. +1! –  J. M. Sep 6 '11 at 4:18
    
Under Another way, you make the claim that the average velocity passing the window is $(v+u)/2$, but average velocity is averaged over distance, not time. It should be time to pass the window divided by the window height=1.35/.21 or about 6.43 m/sec. The time at low speed is weighted higher in this calculation. –  Ross Millikan Sep 6 '11 at 4:20
    
@Ross Millikan: Agreed. But $w=1.35$ and $s=0.210$, so the formula you wrote is the same as the one that I wrote. (The words before that are not the same.) To calculate another way, displacement at $t_1$ is $at_1^2/2$, displacement at $t_2$ is $at_2^2/2$, average velocity is change in displacement divided by change in time. Divide. We get $(at_2+at_1)/2$. But $at_2=v$, $at_1=u$, so indeed average velocity is $(v+u)/2$. –  André Nicolas Sep 6 '11 at 4:50
    
@J.M.: Generality is part of the point. But in addition, I find that students often go to the calculator far too early. Important structural information can get lost in the blizzard of digits. –  André Nicolas Sep 6 '11 at 5:00
    
Yes, I agree with the loss of structural information when one plugs in too early. I've seen kids get lost in manipulations because they consolidated their numbers too quickly. –  J. M. Sep 6 '11 at 5:05
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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 $h(q/g)=\frac{q^2}{2g} \approx 1.48$.

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