Physics, yet again.

Mr. X · 02-13-2010, 08:53 PM

A ball is thrown from the ground onto a roof of height 15 m from a distance of 7 meters away as shown in the diagram. The maximum height of the ball's trajectory is 3.4 meters above the top of the roof.

(a) Find the required initial vertical component of the velocity, Vy.

b) Find the time for the ball to reach maximum height.

(c) Find the time for the ball to fall from the maximum height to the rooftop.

(d) Find the initial horizontal component of the velocity.

Mr. X · 02-13-2010, 08:55 PM

A ball is thrown into the air from ground level. After a time t = 2 s, the ball has traveled to a position x1 = 26 m to the right of and y1 = 9 m up from where it was thrown (at this time, the x and y components of the ball's velocity are still positive). The axes show the x and y directions to be considered positive.

a) What was the x component of the initial velocity of the ball?

13 m/s

b) What was the y component of the initial velocity of the ball?

c) What was the initial speed of the throw?

d) What was the initial angle of the throw relative to the horizontal? Please enter your answer in degrees.

e) What is the height of the ball at the top of its path?

I should get it if I know what b is.

Mr. X · 02-13-2010, 08:55 PM

A ball is thrown from the brink of a cliff 11 m high with an initial velocity of v0 = 4 m/s at an angle q0 = 31° above the horizontal.

a) How long does it take to land?

b) How far from the base of the cliff does it land?

c) What is its maximum height above the base of the cliff?

Mr. X · 02-13-2010, 08:58 PM

At t = 0, an object is projected with a speed v0 = 35 m/s at an angle q0 = 21.7° above the horizontal. The axes on the diagram show the x and y directions that are to be considered positive.

a) The vertical acceleration of the object:

-9.8 m/s^2

b) Its horizontal acceleration:

0

Its vertical velocity:

-85.06 m/s

d) Its horizontal velocity:

32.74 m/s

e) The angle to the horizontal at which the object is traveling (an angle above the horizontal should be reported as a positive number; an angle below the horizontal should be reported as a negative number). Please give your answer in degrees:

f) Its vertical displacement, from where it started

g) Its horizontal displacement, from where it started

h) At what time does the object reach its maximum height?

Figured it out until E....

Mr. X · 02-13-2010, 09:01 PM

river 660 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.

a) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?

d = 1320ft

b) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight accross, as in part (a)), how far downstream is she swept before reaching the opposite bank?

c) For the conditions of part (b), how long does it take for her to reach the opposite bank?

206.57 seconds

I don't know how to get b, I figured out the time for c, but can't get b!!! It's annoying me.

6to7to8 · 02-13-2010, 09:03 PM

Stop posting your homework for other people to do, you cheater.

Mr. X · 02-13-2010, 09:13 PM

That's about 1/4th of the homework, these points are nearly insignificant (5pts per homework, despite the hours that need to be spent doing them). I want to know how to do the damn problems.

-kthxbye.

James has AIDS · 02-13-2010, 09:28 PM

Quote:

Originally Posted by Mr. X

A ball is thrown from the ground onto a roof of height 15 m from a distance of 7 meters away as shown in the diagram. The maximum height of the ball's trajectory is 3.4 meters above the top of the roof.

(a) Find the required initial vertical component of the velocity, Vy.

vf^2 = vi^2 + 2ad

0 = vi^2 + 2 * -9.8 * (15 + 3.4)

vi = 18.99 m/s

b) Find the time for the ball to reach maximum height.

d = vi(t) + .5(a)(t^2)
18.4 = 18.99(t) + .5(-9.8)(t^2)
t = 1.952 s

(c) Find the time for the ball to fall from the maximum height to the rooftop.
d = vi(t) + .5at^2
-3.4 = 0(t) .5(-9.8)(t^2)
t = .833 s
(d) Find the initial horizontal component of the velocity.

I have no idea how to do this. It doesn't seem like there is enough information.

I always have only one variable out of three.

The only equation you can use to solve this is

x distance = time * x velocity

James has AIDS · 02-13-2010, 09:33 PM

Quote:

Originally Posted by Mr. X

A ball is thrown into the air from ground level. After a time t = 2 s, the ball has traveled to a position x1 = 26 m to the right of and y1 = 9 m up from where it was thrown (at this time, the x and y components of the ball's velocity are still positive). The axes show the x and y directions to be considered positive.

a) What was the x component of the initial velocity of the ball?

13 m/s

b) What was the y component of the initial velocity of the ball?

9 = vi(2) + .5(-9.8)(2)^2

vi = 14.3 m/s

c) What was the initial speed of the throw?

d) What was the initial angle of the throw relative to the horizontal? Please enter your answer in degrees.

e) What is the height of the ball at the top of its path?

I should get it if I know what b is.

I can do this rest if you want.

James has AIDS · 02-13-2010, 09:40 PM

Are you doing vector problems?

Quote:

Originally Posted by Mr. X

A ball is thrown from the brink of a cliff 11 m high with an initial velocity of v0 = 4 m/s at an angle q0 = 31° above the horizontal.

a) How long does it take to land?

d = vit + .5at^2

Because angles are involved, vi must be 4sin(31) = 2.06 m/s

-11 = 2.06t + .5(-9.8)t^2

t = 1.723 s

b) How far from the base of the cliff does it land?
xdistance = t * xvelocity

xdistance = 1.723 * 4cos(31)

xdistance = 5.908 m

c) What is its maximum height above the base of the cliff?
When it is at a max, vf will be 0

vf^2 = vi^2 + 2ad

0 = 2.06^2 + 2(-9.8)(x)
x = .217 m above the cliff

Not too sure about that one.

ddsf

James has AIDS · 02-13-2010, 09:44 PM

Quote:

Originally Posted by Mr. X

At t = 0, an object is projected with a speed v0 = 35 m/s at an angle q0 = 21.7° above the horizontal. The axes on the diagram show the x and y directions that are to be considered positive.

a) The vertical acceleration of the object:

-9.8 m/s^2

b) Its horizontal acceleration:

0

Its vertical velocity:

-85.06 m/s

d) Its horizontal velocity:

32.74 m/s

e) The angle to the horizontal at which the object is traveling (an angle above the horizontal should be reported as a positive number; an angle below the horizontal should be reported as a negative number). Please give your answer in degrees:

f) Its vertical displacement, from where it started

g) Its horizontal displacement, from where it started

h) At what time does the object reach its maximum height?

Figured it out until E....

I don't think b and c are right.

When you have angles it, becomes a vector problem.

x velocity becomes initial velocity * cos(angle above horizontal)

y velocity becomes initial velocity * sin(angle above horizontal)

Is this familiar to you?

Think about it: An angle of 21.7 degrees above the horizontal means it is going up, so how is either velocity negative?

Mr. X · 02-13-2010, 09:51 PM

Quote:

Originally Posted by Slanket

I have no idea how to do this. It doesn't seem like there is enough information.

I always have only one variable out of three.

The only equation you can use to solve this is

x distance = time * x velocity

I'll work on it.

Quote:

Originally Posted by Slanket

I can do this rest if you want.

I'll let you know if I don't get it.

Quote:

Originally Posted by Slanket

Are you doing vector problems?

ddsf

Yes, vectors

The one you weren't sure about was wrong.

Quote:

Originally Posted by Slanket

I don't think b and c are right.

When you have angles it, becomes a vector problem.

x velocity becomes initial velocity * cos(angle above horizontal)

y velocity becomes initial velocity * sin(angle above horizontal)

Is this familiar to you?

Think about it: An angle of 21.7 degrees above the horizontal means it is going up, so how is either velocity negative?

I did a bunch of problems like that, but the system said it was right. (remember it's online and tells me when the problem is right)

James has AIDS · 02-13-2010, 09:53 PM

Quote:

Originally Posted by Mr. X

river 660 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.

a) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?

d = 1320ft

b) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight accross, as in part (a)), how far downstream is she swept before reaching the opposite bank?

I think this is just:

tan(37) = d/660

d = 497.346

But I'm not sure
c) For the conditions of part (b), how long does it take for her to reach the opposite bank?

206.57 seconds

I don't know how to get b, I figured out the time for c, but can't get b!!! It's annoying me.

These are relative velocity problems. I'm not great at them.

James has AIDS · 02-13-2010, 09:55 PM

Quote:

Originally Posted by Mr. X

I'll work on it.

I'll let you know if I don't get it.

Yes, vectors

The one you weren't sure about was wrong.

I did a bunch of problems like that, but the system said it was right. (remember it's online and tells me when the problem is right)

Sorry. I had one of those on my midterm, and I didn't know how to do it so I just used calculus lol.

Mr. X · 02-13-2010, 09:58 PM

Quote:

Originally Posted by Slanket

These are relative velocity problems. I'm not great at them.

Nope, I'll keep working on that, probably tomorrow I'm fried right now. I'll do some Psych so I don't really have to think and I can save time tomorrow.

Quote:

Originally Posted by Slanket

Sorry. I had one of those on my midterm, and I didn't know how to do it so I just used calculus lol.

haha, yea it's all good.

James has AIDS · 02-13-2010, 10:00 PM

Just out of curiosity, if you get it wrong, can you put in another answer?

If yes, I will feel a lot better about wrong answers.

Mr. X · 02-13-2010, 10:03 PM

Yes, I can put it as many times as I want. Thankfully, or else this shit better be worth more than 5 points.

Mr. X · 02-14-2010, 11:09 AM

Quote:

Originally Posted by Slanket

I don't think b and c are right.

When you have angles it, becomes a vector problem.

x velocity becomes initial velocity * cos(angle above horizontal)

y velocity becomes initial velocity * sin(angle above horizontal)

Is this familiar to you?

Think about it: An angle of 21.7 degrees above the horizontal means it is going up, so how is either velocity negative?

Still can't get this one...

Or that one with the swimmer's angle.

02-13-2010, 08:53 PM	#1 (permalink)
Mr. X Mr. X is The Beefcake Soulreaper Join Date: Sep 2007 Posts: 24,432 164233 :	Physics, yet again. A ball is thrown from the ground onto a roof of height 15 m from a distance of 7 meters away as shown in the diagram. The maximum height of the ball's trajectory is 3.4 meters above the top of the roof. (a) Find the required initial vertical component of the velocity, Vy. b) Find the time for the ball to reach maximum height. (c) Find the time for the ball to fall from the maximum height to the rooftop. (d) Find the initial horizontal component of the velocity. __________________ Inception of a Grand Crusade (my workout journal) ISSA Certified Fitness Trainer Current Stats: Body Weight: 201lbs Deadlift: 535lbs Power Squat: 462.5lbs CoC #2.5x2 HG300x5 Hub Lift: 45lbs x 2 Rolling Thunder: 150lbs One Arm Barbell Snatch: 125x1 Dumbbell Clean and Jerk 125x4 NGMF 9thG8 Syndicate Stats are from September 09, now I just cry when I look at it. These days I just live in self-pity and don't workout.

02-13-2010, 08:55 PM	#2 (permalink)
Mr. X Mr. X is The Beefcake Soulreaper Join Date: Sep 2007 Posts: 24,432 164233 :	A ball is thrown into the air from ground level. After a time t = 2 s, the ball has traveled to a position x1 = 26 m to the right of and y1 = 9 m up from where it was thrown (at this time, the x and y components of the ball's velocity are still positive). The axes show the x and y directions to be considered positive. a) What was the x component of the initial velocity of the ball? 13 m/s b) What was the y component of the initial velocity of the ball? c) What was the initial speed of the throw? d) What was the initial angle of the throw relative to the horizontal? Please enter your answer in degrees. e) What is the height of the ball at the top of its path? I should get it if I know what b is. __________________ Inception of a Grand Crusade (my workout journal) ISSA Certified Fitness Trainer Current Stats: Body Weight: 201lbs Deadlift: 535lbs Power Squat: 462.5lbs CoC #2.5x2 HG300x5 Hub Lift: 45lbs x 2 Rolling Thunder: 150lbs One Arm Barbell Snatch: 125x1 Dumbbell Clean and Jerk 125x4 NGMF 9thG8 Syndicate Stats are from September 09, now I just cry when I look at it. These days I just live in self-pity and don't workout.

02-13-2010, 08:55 PM	#3 (permalink)
Mr. X Mr. X is The Beefcake Soulreaper Join Date: Sep 2007 Posts: 24,432 164233 :	A ball is thrown from the brink of a cliff 11 m high with an initial velocity of v0 = 4 m/s at an angle q0 = 31° above the horizontal. a) How long does it take to land? b) How far from the base of the cliff does it land? c) What is its maximum height above the base of the cliff? __________________ Inception of a Grand Crusade (my workout journal) ISSA Certified Fitness Trainer Current Stats: Body Weight: 201lbs Deadlift: 535lbs Power Squat: 462.5lbs CoC #2.5x2 HG300x5 Hub Lift: 45lbs x 2 Rolling Thunder: 150lbs One Arm Barbell Snatch: 125x1 Dumbbell Clean and Jerk 125x4 NGMF 9thG8 Syndicate Stats are from September 09, now I just cry when I look at it. These days I just live in self-pity and don't workout.

02-13-2010, 08:58 PM	#4 (permalink)
Mr. X Mr. X is The Beefcake Soulreaper Join Date: Sep 2007 Posts: 24,432 164233 :	At t = 0, an object is projected with a speed v0 = 35 m/s at an angle q0 = 21.7° above the horizontal. The axes on the diagram show the x and y directions that are to be considered positive. a) The vertical acceleration of the object: -9.8 m/s^2 b) Its horizontal acceleration: 0 Its vertical velocity: -85.06 m/s d) Its horizontal velocity: 32.74 m/s e) The angle to the horizontal at which the object is traveling (an angle above the horizontal should be reported as a positive number; an angle below the horizontal should be reported as a negative number). Please give your answer in degrees: f) Its vertical displacement, from where it started g) Its horizontal displacement, from where it started h) At what time does the object reach its maximum height? Figured it out until E.... __________________ Inception of a Grand Crusade (my workout journal) ISSA Certified Fitness Trainer Current Stats: Body Weight: 201lbs Deadlift: 535lbs Power Squat: 462.5lbs CoC #2.5x2 HG300x5 Hub Lift: 45lbs x 2 Rolling Thunder: 150lbs One Arm Barbell Snatch: 125x1 Dumbbell Clean and Jerk 125x4 NGMF 9thG8 Syndicate Stats are from September 09, now I just cry when I look at it. These days I just live in self-pity and don't workout.

02-13-2010, 09:01 PM	#5 (permalink)
Mr. X Mr. X is The Beefcake Soulreaper Join Date: Sep 2007 Posts: 24,432 164233 :	river 660 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water. a) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank? d = 1320ft b) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight accross, as in part (a)), how far downstream is she swept before reaching the opposite bank? c) For the conditions of part (b), how long does it take for her to reach the opposite bank? 206.57 seconds I don't know how to get b, I figured out the time for c, but can't get b!!! It's annoying me. __________________ Inception of a Grand Crusade (my workout journal) ISSA Certified Fitness Trainer Current Stats: Body Weight: 201lbs Deadlift: 535lbs Power Squat: 462.5lbs CoC #2.5x2 HG300x5 Hub Lift: 45lbs x 2 Rolling Thunder: 150lbs One Arm Barbell Snatch: 125x1 Dumbbell Clean and Jerk 125x4 NGMF 9thG8 Syndicate Stats are from September 09, now I just cry when I look at it. These days I just live in self-pity and don't workout.

02-13-2010, 09:03 PM	#6 (permalink)
6to7to8 6to7to8 is A Retired Mod ‎‎‎‎‎‎‎ Join Date: Oct 2008 Posts: 2,542 13740 :	Stop posting your homework for other people to do, you cheater. __________________ ATM 5000 Gangbang Champion

02-13-2010, 09:13 PM	#7 (permalink)
Mr. X Mr. X is The Beefcake Soulreaper Join Date: Sep 2007 Posts: 24,432 164233 :	That's about 1/4th of the homework, these points are nearly insignificant (5pts per homework, despite the hours that need to be spent doing them). I want to know how to do the damn problems. -kthxbye. __________________ Inception of a Grand Crusade (my workout journal) ISSA Certified Fitness Trainer Current Stats: Body Weight: 201lbs Deadlift: 535lbs Power Squat: 462.5lbs CoC #2.5x2 HG300x5 Hub Lift: 45lbs x 2 Rolling Thunder: 150lbs One Arm Barbell Snatch: 125x1 Dumbbell Clean and Jerk 125x4 NGMF 9thG8 Syndicate Stats are from September 09, now I just cry when I look at it. These days I just live in self-pity and don't workout.

02-13-2010, 10:00 PM	#16 (permalink)
James has AIDS James has AIDS is HIV+ Friday bathes in tears Assault Paintball Champion! Join Date: Nov 2007 Posts: 8,849 39988 :	Just out of curiosity, if you get it wrong, can you put in another answer? If yes, I will feel a lot better about wrong answers. __________________ Those who carry a knife and a pear are never afraid of the dark.

02-13-2010, 10:03 PM	#17 (permalink)
Mr. X Mr. X is The Beefcake Soulreaper Join Date: Sep 2007 Posts: 24,432 164233 :	Yes, I can put it as many times as I want. Thankfully, or else this shit better be worth more than 5 points. __________________ Inception of a Grand Crusade (my workout journal) ISSA Certified Fitness Trainer Current Stats: Body Weight: 201lbs Deadlift: 535lbs Power Squat: 462.5lbs CoC #2.5x2 HG300x5 Hub Lift: 45lbs x 2 Rolling Thunder: 150lbs One Arm Barbell Snatch: 125x1 Dumbbell Clean and Jerk 125x4 NGMF 9thG8 Syndicate Stats are from September 09, now I just cry when I look at it. These days I just live in self-pity and don't workout.

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