Calculating the Height of a Window Using Kinematic Equations

[Sorry!]

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
$$d=V_{1}t\frac{at^2}{2}$$
then just rearrange end up with t=$$\sqrt{\frac{2d}{a}}$$

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

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 :|

 

[Doc Al]

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

Good.

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

Redo the substitution more carefully. The converter equation is:
$$d = v_0t + 1/2at^2$$

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

 

[Moetasim]

As a scientist, it is important to always double check your calculations and your assumptions. In this case, it seems like you may have made a mistake in setting up your equations.

Firstly, when listing the given values for the TV, you have listed the initial velocity (v1) as 0 and the final velocity (v2) as x. However, in this scenario, the TV is being dropped, so the initial velocity should be x (since it is moving downwards) and the final velocity should be 0 (since it hits the ground and stops). This may have affected your subsequent calculations.

Additionally, when using the equation d=V1t+(at^2)/2, you have used the value of 9.8 for acceleration (a). However, this value is for the acceleration due to gravity on Earth, and in this scenario, the TV is being dropped from a window, which may not necessarily be on Earth. It is important to always use the correct value for acceleration in your calculations.

Another possible mistake is in your substitution of values into the converter equation. It is not clear how you have arrived at the equation d=(20√(2d/a)+1/2a(2d/a))/2. This equation does not match the original equation d=V1t+(at^2)/2, and it is not clear how you have substituted the values into this equation.

In conclusion, it is important to carefully review your calculations and assumptions to ensure accuracy. Double check your equations and make sure you are using the correct values for the given scenario. It is also helpful to have someone else review your work to catch any mistakes or errors that you may have missed.