Conceptual Physics: Solving Problems in Math & Science

[proculation]

Hummm.. as I reread your question, it would be it be if I was the teacher I would ask for the height of the table.

 

[Dsoto489]

yea the problem is looking for the height of the shelf/table

 

[proculation]

you again mixed the X and Y.

1.92m is where the ball will hit the ground from the shelf, horizontally.

 

[Dsoto489]

ok so the variables i have are

Vo=8m/s
a=0
t=.24s
d=1.92m

a=9.81m/s/s

 

[proculation]

Gimme some minutes, I will draw something.

 

[Dsoto489]

I won't be leaving anytime soon, so take all the time you need. lol thanks.

 

[proculation]

There.

 

[proculation]

See the big black line between X and Y ? That's on purpose. They are to be resolved SEPARATELY.

 

[Dsoto489]

k, i ll try out the problem, give me a sec or 2

 

[proculation]

oops there's a error. it's "a = Vf X t² + Vo"

Sorry.

I was trying to make a good drawing lol

 

[proculation]

damn it misses a 0.5 too.

 

[proculation]

Anyway you have the formula. I drawed it very quickly

 

[Dsoto489]

k so i did some computing and ended up with Vo=170.3125

 

[Dsoto489]

and D= 41.15

 

[proculation]

I mixed up, it's my fault. I will correct my drawing gimme some minutes.

 

[Dsoto489]

no worries.

 

[Dsoto489]

ok, so I found Veloc of y =2.35m/s and Veloc of x =8m/s

 

[Dsoto489]

I have 9.79m tall for the height

 

[proculation]

I corrected

(sorry I'm a little drunk, had problem with my integrals lol)

 

[Dsoto489]

its cool,

with that drawing, the D of X is 4m methinks

 

[proculation]

damn forget that one too.

 

[proculation]

Now.

 

[proculation]

Now 2

 

[proculation]

I guess if there are people watching this they are laughing haha.

 

[Dsoto489]

if i get a 0 in between the parentheses can i multiply it by 1/2

 

[proculation]

if i get a 0 in between the parentheses can i multiply it by 1/2

zero times 0.5 = zero.

That's why I corrected my mistake... I forgot the K of the integral

 

[Dsoto489]

sweet

 

[Dsoto489]

d=.06780672

 

[Dsoto489]

Thank you for the help, i will try again tomorrow.

 

[proculation]

AAh you really want me to get my pencils out to resolve that eh ? :P

 

[Dsoto489]

Ok! So after reworking the problem with different numbers this is what i did and the problem i needed to resolve:
A ball rolls off a shelf which is 2.8 meters high. If it leaves the shelf with a horizontal velocity of 6 m/s, calculate the distance from the base of the shelf at which the ball will strike the ground.

first i used Dvert=1/2at^2 so, 2.8=1/2 x 9.81 x t^2 solving for the time- the time it would take the ball to hit the floor. after square rooting it took aprox .75s to fall.

using Dhoriz=Vot D=6 x .75 the distance from the base at which the ball will strike the ground is 4.53m!

-There is only one other problem I am unable to resolve...

taking a trip you travel 63mph for 30 min, 22mph for 45 min and stop to eat for 20 min. What is your average speed for the entire trip?

what i keep doing is adding the speeds and averaging them, 63+22+0/3=28.33, this answer is wrong(the computer program for school i am using keeps telling me this) Total time is 95 min=1.55hours, i have tried several different ways and just do not get this problem(it seems easy though!)edit+ 85/1.55 =54.83

If you or someone could show me how to approach these types of problems dealing with different speeds over time that would be great. Thanks again proc!