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

[kLPantera]

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=V_ot + (1/2)at²
y=V_ot + (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 = ?
V_o = 22
a_x = 0
a_y = -9.8m/s²
t = ?

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

y = V_ot + (1/2)at²
y = (22)(1.6) + (1/2)(-9.8)(1.6)²
y = 35.2 - 12.544
y = 22.656
Thanks

 

[Dramen]

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

 

[kLPantera]

I don't know if my calculation for time is right though.

 

[zgozvrm]

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).

 

[kLPantera]

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 Vi_y I should do:
Vi_x = V_o cos(theta)?

 

[zgozvrm]

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.

So to find Vi_y I should do:
Vi_x = V_o cos(theta)?

No...
To find Vi_y, you use Vi_y = V_o sin(theta).

Vi_x = V_o cos(theta) is obviously used to find Vi_x.

 

[kLPantera]

V_o = 22 m/s as well?

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

I tried an equation:
V_fy = Vi_y+a_yt
0 = Vi_y + (-9.8)(1.6)
Vi_y = 15.68 m/s

Delta y = Vi_yt + (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?

 

[zgozvrm]

Delta y = 12.544 meters

Is this correct?

Your answer is correct, but your technique is still wrong...

V_fy = Vi_y+a_yt
0 = Vi_y + (-9.8)(1.6)

This says that after falling some distance, the velocity of the ball is Vi_y = 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.

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

Yes, but it's given to you...
"A ball is thrown horizontally..."

 

[kLPantera]

Oh right how did I miss that...

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

So then Vi_y = 0

Then use the equation:

Delta y = Vi_yt + (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

V_fy² = V_iy² + 2a_y Delta y
0 = V_iy² +2(-9.8)(56)
V_iy² = 1097.6
V_iy = 33.13

V_iy = V_osin(theta)
33.13 = V_osin(-51.215)
V_o = 33.13/sin(-51.215)
V_o = -25.8249

 

[zgozvrm]

Oh right how did I miss that...

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

So then Vi_y = 0

Then use the equation:

Delta y = Vi_yt + (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).

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

I did

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

V_fy² = V_iy² + 2a_y Delta y
0 = V_iy² +2(-9.8)(56)
V_iy² = 1097.6
V_iy = 33.13

V_iy = V_osin(theta)
33.13 = V_osin(-51.215)
V_o = 33.13/sin(-51.215)
V_o = -25.8249

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

Again: "A ball is thrown horizontally..."

 

[kLPantera]

Since the ball is thrown horizontally that means V_o = Vi_x right? Since the Vi_y = 0 because there is no vertical velocity. This also means that (theta) = 0 degrees right?

But when I try to find Vi_y I get:

Vi_y = V_osin(theta)
V_o = 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 Vi_y and then use the equation: Delta y = Vi_yt + (1/2)gt² to find time.

Then use the equation: V = V_o + gt to find V_o.

 

[zgozvrm]

Since the ball is thrown horizontally that means V_o = Vi_x right? Since the Vi_y = 0 because there is no vertical velocity. This also means that (theta) = 0 degrees right?

But when I try to find Vi_y I get:

Vi_y = V_osin(theta)
V_o = 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.<br /> <br /> <br /> You'll have to go about it a different way...[/itex][/itex]

 

[kLPantera]

Is it plausible to first find time then use the equation:

Delta x = Vi_xt + (1/2)at²

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

So I could find Vi_x = Delta x/t

And I think Vi_x = V_o, that would be my answer?

 

[zgozvrm]

You got it!

 

[zgozvrm]

... the problem is: you only know [itex]\Delta[/tex]x<br /> so you need to find either Vi_x or t ...[/itex]

 

[kLPantera]

I tried a different way by doing:

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

Then:

Delta x = V_ot + (1/2)a_xt²
45 = Vi_x(3.3)
Vi_x = 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.

 

[zgozvrm]

Great job!

 

[zgozvrm]

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...