5 Dollars On Paypal

I got the answer all i gotta do is show the work. 5 dollars on paypal for whoever gives me the work!
Grade 11 physics
A small object is floating down a river at a constant velocity of 1.0m/s a student stand on a bridge drops a stone when the object is 4.9m upstream from the bridge. Ifthe stone hits the object, from what height was the object dropped.
Answer is 78 m


A stone is thrown vertically downward from the top of a 96m building at a velocity of 5.0m/s this stone is observerd passing by a window that has a height of 2.0m. Assuming that the bottom of the window is 25m from the ground how long time does it take for the stone to pass by the window?
Answer: 5.3 x 10^-2 s

A student throws a stone vertically upward from the base of a 51m building at the same time a second student on the top of a the building drops a stone. If the two stones collide 12 m from the ground what was the initial velocity of the stone thrown.
Answer: 18m/s

All i need is to show work!!

5 dollars on paypal!!

PM the work!

 

When you need this by?


Can you pm me any formulas?

 

me too? I take physics at 9th grade, too advanced for j00

 

I am in 12th grade
and this is really easy work
dunno why you dont get it :\

anyways I aint got the time for it
or I would help you bro

Zain


pS: good luck finding one who can do em

 

here are the formulas
d = vit-1/2at^2
a= vf-vi/t
d=((vf-Vi)/2)t
d= vit+1/2at^2
Vf2=Vi^2+2ad

 

rofl..I did this in 9th grade. All you do is plug in numbers and stuff. I would make these kinds of offers if i had the money. Buncha grunt work . Bump for ya

 

your answe for the 1st question look fishy ...
here are my answers with workings

Qn1
Time taken by the object to travel
before it was hit=4.9m/1m/s=4.9s
hight=1/2*g*t^2
g=9.8m/s^2;t=4.9s
ht=9.8*4.9*4.9/2=117.649m

Q2
u=initial velocity=5.0m/s
a=9.81m/s^2
s=69m
v=?

v^2 = u^2 + 2as
v^2 = 5^2 + 2*9.81*69
v=37.13

u1=37.13m/s
a=9.81m/s^2
s=2m
t=?

s = ut + 1/2at^2
2=37.13*t + 1/2*9.81*t^2
4.905t^2 + 37.13t - 2 = 0
t = 0.053s


Q3
u(top) = 0
a = 9.81
s = 51 - 12 = 39
t = ?

s = ut + 1/2at^2
39 = 0 + 1/2*9.81*t^2
t = 2.8198 (collision time)

t(bottom) = 2.8198
a = -9.81
s(bottom) = 12
u(bottom) = ?

s = ut + 1/2at^2
12 = u*2.8198 + 1/2*(-9.81)*2.8198^2
u = 18.0m/s


$5 to my paypal at [email protected]

thanks :lol:

 

wtf ? i dont understand a thing .. sorry i spam but you guys are great haha .

 

congrats tazzycorner you earned that 5$. On the other hand it was smple formulas ^_^

 

lol

everyone seems to think it's easy huh?

too lazy to type the answers out I suppose :lol:

 

you were faster, you already answered this one before i noticed this topic ;(

 

damn .. didn't notice this before
this was easy ... tazzy gj

I did those things 3 years ago
it's not that hard

 

lol ... thanks

now, let's see if I get the $ :lol:

 

GFX Spawn : u saw my PM 2 hours ago ...

how are the answers?

do drop me a reply

I will be waiting