# Thread: Quadratic Forumla: Dropping an object

1. ## Quadratic Forumla: Dropping an object

A bit of a repost, but I have a different question. This is the question given:

If an object is thrown downward at 26 m/s from the top of a building, 102m high, the equation relating to displacement (meters) and time (seconds) is:

$\displaystyle ]s(t) = 1/2 gt^2 + 26t + 102$

Find the time for an object to reach the ground from the top. Use g=-9.81

So I have come up with

$\displaystyle (-26) +/- \sqrt (26)^2 - 4(-4.905)(102) / 2(-4.905)$

Unfortunately I keep getting 2 negative solutions.

I was given this explaination:

The initial velocity is $-26 {\text{ m/s}$ (you are working with s positive upwards) and $g=-9.81 {\text{ m/s}}^2$, so your equation for displacement is:

$s(t)=- \frac{9.81t^2}{2}-26t+102$

which I can tell you has exactly one positive root (and in consequence one negative root, but that is not the one you want).

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My question is, how do I figure out I need to use -26t? Am I supposed to look at the given formula and change 26t to a negative number because I am dropping an object?

2. ## Re: Quadratic Forumla: Dropping an object

Originally Posted by polskon

My question is, how do I figure out I need to use -26t? Am I supposed to look at the given formula and change 26t to a negative number because I am dropping an object?
It depends on how you choose your frame of reference ...
In this case they have chosen the 'y-axis' upwards and because your acceleration and velocity are orientated downwards (in this case) they get a negative sign.

3. ## Re: Quadratic Forumla: Dropping an object

Id just like to update: My problem with the formula was that I kept getting 2 negative solutions. To my surprise, it was my fault. I was continuously inputting the answers wrong on my calculator. I guess I didn't understand the formula 100%. Thanks!