Calculating Initial Velocity of a Thrown Object Using Free-fall Equations

[HRubss]

Homework Statement

A man stands on the edge of a roof 55m above the ground, he throws a stone upward and drops a second stone at the same time. If the dropped stone hits the ground 4.4s before the stone that was thrown upward, what was the initial velocity of the first one?

Homework Equations

1.) Δy = v₀²t + ½(a)t²
2.) v_i = v₀ + at
3.) (v_i)² = (v₀)² +2a(Δy)

The Attempt at a Solution

I just wanted to check if my work is correct. First thing I did was realize that the first stone has two different time frame, first time frame being the point at which it is released from the mans hand which has the initial velocity I need, and the 2nd time frame in which the stone is in mid air about to be in free-fall so it would have a velocity of 0m/s. Most of the information is known about the 2nd stone. I made a quick list of variables known.

stone 1 pt 1 (thrown up-point at which it's about to be in free-fall)
y₀ = 55m
y_i = ?
a = -9.81m/s²
v_i = 0m/s
v₀ = ?

stone 1 pt 2 (free-fall-hits the ground)
y₀ = ?
y_i = -55m
v₀ = 0m/s
a = -9.81m/s
t = ? (+4.4s)

stone 2
y₀ = 55m
y_i = 0m
v₀ = 0m/s
a = -9.81m/s²
t = ?

I realized that the thing that connected pt 2 of stone 1 and stone 2 is the time it took to hit the ground so I used equation 1 to find t and it gave me 3.34s
I added that 3.34s to 4.4s to get 7.74s since it took 4.4s longer for the 2nd stone to hit the ground. Using 7.74s, I used equation 2 to find the height of which stone 1 at pt 2 was and I got 238m (I used equation 1 to check my answer and got the same thing). Now that I had the height of pt 2 to stone one, I plugged it in equation 3 and got v₀ ≈ 60 m/s for the initial velocity of pt 1. I checked my work and plugged that number into the same equation to find v_i or the velocity at which the stone is about to go into free-fall and I got 0 m/s. I just wanted to confirm if my answers and steps were correct, thanks!

 

[lewando]

A couple things look incorrect beginning with:

1.) Δy = v02 + ½(a)t2

And this looks like stone₁ is ending up underground.

stone 1 pt 2 (free-fall-hits the ground)
y0 = ?
yi = -55m
v0 = 0m/s
a = -9.81m/s
t = ? (+4.4s

 

[HRubss]

And this looks like stone1 is ending up underground.

Whoops, forgot to put in t after initial velocity, I included it in my calculations

A couple things look incorrect beginning with:

I just realized that if the stone was thrown upward, it wouldn't travel 55m down because the roof itself is 55m above the ground. How would I go about finding the correct position?

EDIT:
Instead of making y₀ = 0m and y_i = 55m (from mid-air free-fall to ground in the pt2 of stone 1, a mistake of mine) I just found Δy during stone 1's free fall and that gave me 293m meaning that the difference in height between the stone 1 at the roof to pt2 of it in free-fall is 293m(Δy)+55m(height of the roof) = 348m?

 

[lewando]

For the "stone 1 pt 2 (free-fall-hits the ground)" setup, you are looking for y_i, the final position (BTW, I don't know why you wouldn't call it y_f -- the final position. Looks too much like y_initial. But don't change anything now). So somewhere in your work you should have a drawing of the situation along with a y-axis origin identified somewhere. From your "stone 1 pt 1 (thrown up-point at which it's about to be in free-fall)" setup, it looked like you have chosen the surface of the Earth to be the y-axis origin. If this is the case, then don't you know the position of the final resting place of either stone?

 

[HRubss]

For the "stone 1 pt 2 (free-fall-hits the ground)" setup, you are looking for yi, the final position (BTW, I don't know why you wouldn't call it yf -- the final position. Looks too much like yinitial.

Yeah it messes me up too sometimes since everyone else uses vf but my professor uses v1 and v0 instead so that's where I got the habit. I edited post #3.

 

[lewando]

Well v₁ or y₁ would be fine just use the 1 key instead of the i key.

 

[lewando]

Precision in thought and expression is important in solving these kinds of problems. Any problem really.

1.) Δy = v₀²t + ½(a)t2

Is still not quite right.

And the definition of Δy, displacement, is important to get right so as to not have a sign problem.

 

[HRubss]

Precision in thought and expression is important in solving these kinds of problems. Any problem really.

Is still not quite right.

And the definition of Δy, displacement, is important to get right so as to not have a sign problem.

y_f-y_i = v_initial(t)+½at²?

 

[lewando]

Very correct!

 

[HRubss]

Very correct!

What about it? I applied it to my calculations and got an initial velocity of ≈ 83m/s? That doesn't sound right to me..

 

[lewando]

It is not. You need to begin to show your steps so we can find out where the error is coming from.

 

[HRubss]

It is not. You need to begin to show your steps so we can find out where the error is coming from.

I apologize, I will try to upload a photo of my work with the steps when I get the chance sometime soon.

 

[HRubss]

step numbers are shown by pink pen, my work is all over the place lol... please let me know if it's too messy to understand and I'll rewrite everything more organized!

 

[Ray Vickson]

I apologize, I will try to upload a photo of my work with the steps when I get the chance sometime soon.

No, No, No: don't do that! Type it out (as per the PF standards).

 

[HRubss]

Okay, so he's what I did starting from the beginning..
I used 2 reference points for stone 1, one at the moment the stone is throwing upward with an initial velocity I need to find and the second one being at the point the stone reaches it's highest maximum height and falls to the ground. Stone 2 was pretty straight forward..
stone 2 variables:
-y₀ = 55m (height of roof)
-y₁ = 0m (hitting the ground)
-v₀ = 0 m/s
-a = -9.81m/s²
t = ?

I assumed it's important for me to find t because it's really the only thing that links these 3 pieces of information together.
I found t by using (y₁ - y₀) = v₀(t) + ½(a)(t)²
now inserting the numbers into the variables... (0m-55m) = (0m/s)t+½(-9.81m/s²)(t)²
Doing the basic arithmetic, I got t = 3.34s which means it took the 2nd stone that long to hit the ground from start to finish. What I'm realizing now that I goofed up is that I added the 3.34s to the stone 1's time at the point in which it's in free-fall (since it states that stone 2 hit the ground 4.4s BEFORE stone 1) I didn't realize that 3.3s + 4.4s = 7.7s is actually the full time it took stone 1 to hit the ground from start to finish, instead of the stone being in mid air. Now that I have a vital piece of information, the time it took for stone 1 to fall to the ground, I can use it to find the initial velocity.

Stone 1 variables:
-y₀ = 55m
-y₁ = 0m
-v₀ = ?
-v₁ = I don't think I need it since I'm going to be using (y₁ - y₀) = v₀(t) + ½(a)(t)² and that doesn't require v₁.. right?

I set up (y₁ - y₀) = v₀(t) + ½(a)(t)² and rearrange it to get me v₀ instead.
The function now looks like (2(y₁ - y₀) - a(t)²)/t = v₀..
with the t in the bottom, I got 2 fractions, (2Δy)/t - (at²)/t but since t is squared, it cancels out with the bottom t.
After plugging in the numbers and doing the math, I got v₀ = 61.25m/s. If this is right, I don't think it was necessary for me to create two different time frames for stone 1.

 

[lewando]

I set up (y1 - y0) = v0(t) + ½(a)(t)² and rearrange it to get me v0 instead.
The function now looks like (2(y1 - y0) - a(t)²)/t = v0

Check your math on this step.

 

[HRubss]

Check your math on this step.

Looking it up online, the proper equation should have been v₀ = (Δy/t) - (at/2). Looks like instead of multiplying the constant of 2 like I did, they just inversely subtracted the +½(-9.81m/s²)(t²). For future experiences, I should just inversely bring over variables that have a constant in front of them like that instead of split them up? Well, after applying my numbers to the correct formula, I got v₀ = 34m/s.

 

[lewando]

Looking it up online, the proper equation should have been v₀ = (Δy/t) - (at/2).

Why would yo do that when you have a perfectly good equation: (y₁ - y₀) = v₀t + ½(a)(t)²? In a real-life testing situation, you will not have that luxury. Anyway, check your math again, 34 m/s is not correct.

(please ignore the attachment, I have no idea how it got there)

 

[lewando]

For future experiences, I should just inversely bring over variables that have a constant in front of them like that instead of split them up?

I am having difficulty decoding this question. Whatever you do, stay within the bounds of standard algebraic manipulation.

 

[HRubss]

Why would yo do that when you have a perfectly good equation: (y1 - y0) = v0t + ½(a)(t)2? In a real-life testing situation, you will not have that luxury. Anyway, check your math again, 34 m/s is not correct.

Whoops, used -30m instead of -55m. It's not that I wanted to look up my answers, I just didn't trust my arithmetic and I wanted to see how the proper formula was achieved. The correct answer should be 30.6m/s or 31 m/s.

 

[lewando]

That looks right.

 

[HRubss]

That looks right.

Thanks for your help, the answer seems so simple to achieve with just some critical thinking. I think I've just been spoon-fed so much to the point that I don't see the simple things anymore!

 

[lewando]

I wanted to see how the proper formula was achieved.

If you are asserting that there is something improper about: (y₁ - y₀) = v₀t + ½(a)(t)², then that is an improper assertion. You get the same result with either equation.

 

[HRubss]

If you are asserting that there is something improper about: (y₁ - y₀) = v₀t + ½(a)(t)², then that is an improper assertion. You get the same result with either equation.

This is very true, I just check both equations and got the same thing. Having the units cancel out helps.