Test Question

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Homework Statement


Ahh just had a test and i thought i did this question correctly as it turns out though most of the other people in the class got a different value than me so i'm just wondering if someone could show me where i went wrong :D
Bascially the question is a person drops a TV from a window and then throws a converter at 20m/s down it takes the converter half the amount of time to reach the ground how high is the window

so i listed the given values for the TV with down as positive:
v1=0 v2=x a=9.8 d=? t=?
and the converter:
v1=20 v2=x a=9.8 d=? t=?

then i used the TV to find an equation for time since v1=0
[tex]d=V_{1}t\frac{at^2}{2}[/tex]
then just rearrange end up with t=[tex]\sqrt{\frac{2d}{a}}[/tex]

so i sub that into the converter equation get...
[tex]d=\frac{20\sqrt{\frac{2d}{a}}+1/2a\frac{2d}{a}}{2}[/tex]

then i subbed my values and got something like 3/4d^2+20.7d=0 (not sure of exact values..)
d(3/4d+20.7)=0
20.7(4)/3=d
27.6=d
so you get d=0 and d=27.6 so its obviously 27.6 but other people got 33.6 or something where did i go wrong :|
 

Answers and Replies

  • #2
Doc Al
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then i used the TV to find an equation for time since v1=0
[tex]d=V_{1}t\frac{at^2}{2}[/tex]
then just rearrange end up with t=[tex]\sqrt{\frac{2d}{a}}[/tex]
Good.

so i sub that into the converter equation get...
[tex]d=\frac{20\sqrt{\frac{2d}{a}}+1/2a\frac{2d}{a}}{2}[/tex]
Redo the substitution more carefully. The converter equation is:
[tex]d = v_0t + 1/2at^2[/tex]

You need to substitute:
[tex]t = (\sqrt{\frac{2d}{a}})/2[/tex]
 

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