Distance traveled by projectile in Free Fall

[fluidity]

Homework Statement

An object falls from rest off a building, with a constant acceleration of 9.8 m/sec^2.
The object travels half the building in its last 1 secs of its fall before hitting the ground. What is the height of the building?

v_0 = 0
a = 9.8

Homework Equations

x = x_0 + v_0 t + (1/2) a t^2

The Attempt at a Solution

I know that the distance traveled by the projectile for the first half of the building is given by
y= (1/2)gt^2

Now that I know that after traveling that half the height of the building it spends another second traveling the other half.

So now 2y=(1/2)g(t+1)^2

Now by arranging the first equation I get t=√(2y/g)
If I substitute that into the second equation for the variable t I get

2y= y + g√(2y/g) + (1/2)g

After this I've attempted way to get at the right answer such as using the quadratic formula but I got wrong answers. The right answer is 57.1 m, however I haven't been successful in getting to that answer.

 

[Nathanael]

Welcome to Physics Forums.

Well done, your equation does give the right answer. You just made a mistake somewhere in solving it.

If I had to guess, I would say that you probably forgot that this particular quadratic equation yields \sqrt{y} so you must square the answer (and then of course multiply by 2 since you defined the building to be a height 2y)


(Quadratic equation is ax^2+bx+c=0 and so in this case you have a\sqrt{y}^2+b\sqrt{y}+c=0 and so x=\sqrt{y} and so you'll have to square the answer you get)

 

[fluidity]

Wow thanks, I didn't look at the quadratic formula in that perspective. My method of finding the answer was different compared to the Mastering Physics solution, but it gave the same exact number. So once again I want to say thanks, I really appreciate the help!

 

[Nathanael]

My method of finding the answer was different compared to the Mastering Physics solution, but it gave the same exact number.

Don't worry, it happens often. I think your method was an excellent one. (The textbook solutions are not always the simplest.)

I want to say thanks, I really appreciate the help!

No problem! But you did most of the work