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!


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

  • $\begingroup$ "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! $\endgroup$ – J. M. is a poor mathematician Sep 6 '11 at 4:18
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Sep 6 '11 at 4:20
  • $\begingroup$ @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 $(v+u)/2$. $\endgroup$ – André Nicolas Sep 6 '11 at 4:50
  • $\begingroup$ @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. $\endgroup$ – André Nicolas Sep 6 '11 at 5:00
  • $\begingroup$ 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. $\endgroup$ – J. M. is a poor mathematician Sep 6 '11 at 5:05

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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