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

 

[HalfLostAndSearching]

Your approach is on the right track, but there are a few issues with your equations and calculations. Here is a clearer and more systematic approach:

First, we can set up the following equations for the two halves of the building:

First half: y = (1/2)gt^2
Second half: y = (1/2)g(t+1)^2

Next, we can set these two equations equal to each other, since the total distance traveled by the object is equal to the height of the building:

(1/2)gt^2 = (1/2)g(t+1)^2

We can simplify this equation by dividing both sides by (1/2)g, which cancels out on both sides:

t^2 = (t+1)^2

Expanding the right side, we get:

t^2 = t^2 + 2t + 1

Subtracting t^2 from both sides, we get:

0 = 2t + 1

Subtracting 1 from both sides, we get:

-1 = 2t

Dividing both sides by 2, we get:

t = -1/2

This is not a valid solution, as time cannot be negative. This means that our original assumption that the object spends 1 second traveling the second half of the building is incorrect. We need to adjust this assumption to make the solution work.

Let's try assuming that the object spends t seconds traveling the second half of the building. This means that it spends (1-t) seconds traveling the first half of the building. We can set up the following equations for each half:

First half: y = (1/2)g(1-t)^2
Second half: y = (1/2)gt^2

Setting these two equations equal to each other, we get:

(1/2)g(1-t)^2 = (1/2)gt^2

Dividing both sides by (1/2)g, we get:

(1-t)^2 = t^2

Expanding the left side, we get:

1 - 2t + t^2 = t^2

Subtracting t^2 from both sides, we get:

1 - 2t = 0

Adding 2t to both sides, we get:

1 = 2t

Dividing both sides by