How do I calculate the height of a bridge using one dimension?

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SUMMARY

The height of the bridge can be calculated using the equation Delta x = v_0 * t + 0.5 * a * t^2, where v_0 is the initial velocity (0 m/s), a is the acceleration due to gravity (-9.8 m/s²), and t is the time of fall (2.3 seconds). The calculation yields a height of 26 meters. The discussion clarified that the angle notation (0=270 degrees) indicates a vertical drop, confirming that the height should be a positive value, thus the final answer remains 26 meters.

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Homework Statement


The question says this: Use one dimension to solve the problem.
A woman drops a rock off a bridge (0=270 degrees-- wasn't sure how to type the line through the zero, sorry!) and it takes 2.3 seconds for the rock to hit the water below. How tall is the bridge?

Homework Equations


Delta x=volt+one half at^2

The Attempt at a Solution


I have this written out:
Delta x=volt+one half at^2
Delta x=(0)*(2.3 sec) plus one half times (-9.8 m/sec^2) times (2.3)^2
Delta x= one half times (-9.8 m/sec^2) times (5.3 sec^2) ((the sec^2 is slashed out))
Delta x=26 m
I then made the 26 negative, as I read it needs to be.

When I was working the problem, I wasn't sure what to do with the 0=270 degrees portion. I'd like to know if I did this correctly, and if not, what I did incorrectly/what the fix for it is. Thank you! :)
 
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fandomgeek_394 said:
I wasn't sure what to do with the 0=270 degrees
Neither do I. What does it mean?
fandomgeek_394 said:
I then made the 26 negative,
Why? It asks how tall the bridge is. That cannot be negative.
fandomgeek_394 said:
5.3 sec
Is that a typo?
 
Yes, it is a typo. And I realized the negative thing a bit ago; at the time when I posted that I had misunderstood something I read, so I have 26m marked down.
 
fandomgeek_394 said:
Yes, it is a typo. And I realized the negative thing a bit ago; at the time when I posted that I had misunderstood something I read, so I have 26m marked down.
Assuming the 270 degrees means horizontal, 26m is right.
 
fandomgeek_394 said:
(0=270 degrees-- wasn't sure how to type the line through the zero, sorry!)

If you click on the ##\Sigma## icon in the edit window top tool bar, a menu of special characters and symbols will be presented from which you can select your zero with a line, otherwise know as the Greek letter theta: θ :smile:
 
haruspex said:
Assuming the 270 degrees means horizontal, 26m is right.
##\theta=270^\circ## is more consistent with the term "dropping" in the exercise text, i.e. straight down ... :smile:
 
BvU said:
##\theta=270^\circ## is more consistent with the term "dropping" in the exercise text, i.e. straight down ... :smile:
Yes, of course!
 
Thanks, everyone. :)
 

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