Are my calculations right? Also, not sure about a formula?

In summary, the equations used in this problem are: Problem #1: Square root (g/2 delta y) * delta xProblem #2: Delta x = 36Delta y = (1/2)g * delta t squared
  • #1
kLPantera
43
0

Homework Statement



Problem #1

A ball is thrown horizontally from the roof of a building 56m tall and lands 45m from the base. What was the ball's initial speed.

Problem #2

A ball thrown horizontally at 22.2m/s from the roof of a building lands 36.0 meters from the base of the building. How high is the building?

Homework Equations



Problem #1

Not sure which equation(s) I need to use. I know I have to find initial speed, but I think I have to find time to find initial speed. So I'm unsure as to how to find time, because time keeps giving me some trouble.

Problem #2

x=Vot + (1/2)at2
y=Vot + (1/2)gt2

The Attempt at a Solution



Problem #1

Don't know which equation(s) to use so, no attempt.

Problem #2 (Could someone please confirm my calculations for this one?)

Delta x = 36
Delta y = ?
Vo = 22
ax = 0
ay = -9.8m/s2
t = ?

x = Vot+ (1/2)at2
36 = 22t
t = 1.6 seconds

y = Vot + (1/2)at2
y = (22)(1.6) + (1/2)(-9.8)(1.6)2
y = 35.2 - 12.544
y = 22.656
Thanks
 
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  • #2
Problem #1:
As far as I can tell the question is related to projectile motion in 2 dimensions while your equations will only work for 1 dimension motion and because I'm new to physics as well I'll do my best to help you out.

A simple equation I found for initial velocity is: Square root (g/2 delta y) * delta x
In which g= gravity, 2 delta y= delta y * 2And Problem #2

Simple Equation for delta y is (1/2)g * delta t squared

And to find delta t it is delta x/initial velocity*cos angle

And angle is 0
 
  • #3
I don't know if my calculation for time is right though.
 
  • #4
Problem #2:

Your answer is incorrect, although your technique is fine.

The question states that the ball is thrown horizontally at a speed of 22 m/s.
Therefore, the initial horizontal velocity is 22 m/s.

What is the initial vertical velocity? (Hint, it's NOT 22 m/s).
 
  • #5
Problem #2

Is the time I calculated wrong too? Or is it just the second part of my calculations that is wrong?

So to find Viy I should do:
Vix = Vo cos(theta)?
 
  • #6
kLPantera said:
Is the time I calculated wrong too? Or is it just the second part of my calculations that is wrong?
The calculated time is correct.



kLPantera said:
So to find Viy I should do:
Vix = Vo cos(theta)?
No...
To find Viy, you use Viy = Vo sin(theta).

Vix = Vo cos(theta) is obviously used to find Vix.
 
  • #7
Vo = 22 m/s as well?

I would have to calculate theta too wouldn't I?

I tried an equation:
Vfy = Viy+ayt
0 = Viy + (-9.8)(1.6)
Viy = 15.68 m/s

Delta y = Viyt + (1/2)gt2
Delta y = (15.68)(1.6) + (1/2)(-9.8)(1.6)2
Delta y = 25.088 - 12.544
Delta y = 12.544 meters

(The answer some other people in my class calculated was 12.8 meters)

Is this correct?
 
  • #8
kLPantera said:
Delta y = 12.544 meters

Is this correct?
Your answer is correct, but your technique is still wrong...


kLPantera said:
Vfy = Viy+ayt
0 = Viy + (-9.8)(1.6)
This says that after falling some distance, the velocity of the ball is Viy = 0 m/s.
This is not likely, since we know that the acceleration of gravity is 9.8 m/s/s.
The ball can only gain speed.


kLPantera said:
I would have to calculate theta too wouldn't I?
Yes, but it's given to you...
"A ball is thrown horizontally..."
 
  • #9
Oh right how did I miss that...

So since it's thrown horizontally, theta = 0 right?

So then Viy = 0

Then use the equation:

Delta y = Viyt + (1/2)gt2
Delta y = (0)(1.6) + (1/2)(-9.8)(1.6)2
Delta y = -12.544

However since it's a building, the answer would be +12.544 right?

----------------------------------------------------------------

Also I tried Problem #1 some more could you point me in a direction?

I did

Theta = tan-1(-56/45) = -51.215 degrees

Vfy2 = Viy2 + 2ay Delta y
0 = Viy2 +2(-9.8)(56)
Viy2 = 1097.6
Viy = 33.13

Viy = Vosin(theta)
33.13 = Vosin(-51.215)
Vo = 33.13/sin(-51.215)
Vo = -25.8249
 
  • #10
kLPantera said:
Oh right how did I miss that...

So since it's thrown horizontally, theta = 0 right?

So then Viy = 0

Then use the equation:

Delta y = Viyt + (1/2)gt2
Delta y = (0)(1.6) + (1/2)(-9.8)(1.6)2
Delta y = -12.544

However since it's a building, the answer would be +12.544 right?

That's correct. The formula finds the displacement in the y-direction which, in this case, is down (hence the negative result).




kLPantera said:
Also I tried Problem #1 some more could you point me in a direction?

I did

Theta = tan-1(-56/45) = -51.215 degrees

Vfy2 = Viy2 + 2ay Delta y
0 = Viy2 +2(-9.8)(56)
Viy2 = 1097.6
Viy = 33.13

Viy = Vosin(theta)
33.13 = Vosin(-51.215)
Vo = 33.13/sin(-51.215)
Vo = -25.8249

You calculated the angle of displacement; you need the angle of trajectory.

Again: "A ball is thrown horizontally..."
 
  • #11
Since the ball is thrown horizontally that means Vo = Vix right? Since the Viy = 0 because there is no vertical velocity. This also means that (theta) = 0 degrees right?

But when I try to find Viy I get:

Viy = Vosin(theta)
Vo = 0/sin(0)

But that's undefined since sin(0) is 0 and you can't divide by 0.

I was figuring I would find Viy and then use the equation: Delta y = Viyt + (1/2)gt2 to find time.

Then use the equation: V = Vo + gt to find Vo.
 
  • #12
kLPantera said:
Since the ball is thrown horizontally that means Vo = Vix right? Since the Viy = 0 because there is no vertical velocity. This also means that (theta) = 0 degrees right?

But when I try to find Viy I get:

Viy = Vosin(theta)
Vo = 0/sin(0)

But that's undefined since sin(0) is 0 and you can't divide by 0.

You can't do that!

Consider X = 2 * 0
X = 0

But X [itex]\div[/tex] 0 [itex]\ne[/tex] 2.


You'll have to go about it a different way...
 
  • #13
Is it plausible to first find time then use the equation:

Delta x = Vixt + (1/2)at2

Since ax = 0, the second half of the equation would be gone then right?

So I could find Vix = Delta x/t

And I think Vix = Vo, that would be my answer?
 
  • #14
You got it!
 
  • #15
... the problem is: you only know [itex]\Delta[/tex]x
so you need to find either Vix or t ...
 
  • #16
I tried a different way by doing:

Delta y = Viyt + (1/2)(g)(t^2)
-56 = -4.9t2
t= 3.3

Then:

Delta x = Vot + (1/2)axt2
45 = Vix(3.3)
Vix = 13.63

Is this correct now? I think it is.
-----------------------------------

Also would you mind if I asked you about another problem, I'm not sure if my answer is correct because. I did not catch the answer my teacher said it should be.
 
  • #17
Great job!
 
  • #18
kLPantera said:
Also would you mind if I asked you about another problem, I'm not sure if my answer is correct because. I did not catch the answer my teacher said it should be.

Sure, but you should either start a new thread, or PM me...
 

1. Are my calculations accurate?

This is a common question when it comes to scientific calculations. The best way to ensure accuracy is to double check your work and use reliable sources or formulas. It may also be helpful to have a colleague review your work for any potential errors.

2. How do I know if I am using the correct formula?

It is important to carefully read and understand the problem or experiment you are working on. Make sure to choose the appropriate formula based on the given variables and data. If you are unsure, consult a textbook or ask a fellow scientist for clarification.

3. What should I do if my calculations do not match the expected results?

If your calculations do not align with the expected results, double check your work for any errors. It may also be helpful to review the formula and make sure you are using the correct units. If the issue persists, seek advice from a mentor or conduct further research on the topic.

4. Is there a way to check my calculations without redoing them?

Yes, there are various tools and resources available for checking scientific calculations. You can use online calculators, software programs, or even consult a trusted colleague or mentor for a second opinion. It is always a good idea to have someone else review your work to ensure accuracy.

5. Are there any common mistakes to watch out for when doing calculations?

Yes, some common mistakes when doing calculations include mixing up units, using the wrong formula, and not properly rounding numbers. It is important to be meticulous and pay attention to detail when performing calculations to avoid these errors. Also, always double check your work before finalizing your results.

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