/sci/ - Science & Math

You prolly have to assume earth drag. Maybe.

>>5581779
Not at this level of physics.
I guess that you're meant to assume g as -9.8, but even so, I can't seem to get it.

Write down your known values, you have (t=10.2) (u=0) and (x=750) where t is time u is initial velocity and x is distance. Therefore you already know the initial velocity is 0. Next question coming up.

>>5581809
u is 0, yet it flies into the air?

>>5581814
It most likely didn't begin with a running start.

>>5581773
v= d/t
v=t*a
a=v/t
distance post burn=(total velocity-g*t^2)
solve for t, and take integral from a*t_thrust^2 to 0

then you can set that equal to 10.2 seconds.

then find the maximum of the function and subtract x-1.5

you can then solve for distance.

use piecewise functions such as ((((((x-0.5)/sqrt((x-0.5)*(x-0.5)))/2)+0.5)*A)+(((((x-0.5)/sqrt((x-0.5)*(x-0.5)))/-2)+0.5)*B))


trust me I played kerbal space programme.

>>5581820
Ignore me completely.
When it hits the cloud we know a=-9.8, t=10.2 and x= 750
We use x = ut + 1/2(at^2)
Plug in values and we get 750 = 10.2u+1/2(-9.8 x 10.2^2)

This simplifies down to u = 123.51

>>5581827
>>5581773

it's a beautiful problem op and is a perfect example of geometric abstraction in physics.

>>5581828
Now we need to find the time. At the height of the arc it will reach v=0. So we can say that v=u+at
Plug in values again we get
0 = 123.51 + 9.8t
Which works out to be t = 12.6

>>5581833
Finally we find the height by saying x=1/2u+v)t
Plug in values yadda yadda we get x = 778.113 m

>>5581835
1/2(u+v)t*

>>5581836
Finally the thickness of the cloud layer
778.113 - 750 = 28.113 meters

>28.113 meters

don't think so
what happened to the 1.5 seconds?

>>5582620
OP here again, sorry, but this. I assumed that the probe comes back down. It just makes no sense. In any way you interpret it, there is at least one question you can't do. Using the stuff I got a total height of less than 750, which doesn't work

>>5581827
but I don't know about that. I don't have a high enough level of physics.

What a weird problem. Does the probe go through the cloud layer or is it's peak in side the clouds? If so how can you possibly know the thickness if it stops in the middle?

>>5583478
its*

a)
s = 750
u = ?
v = ?
a = -9.8
t = 10.2

calculate u as 750 = 10.2 u - 1/2 9.8 10.2^2

also v for part c from 750 = 10.2 v + 1/2 9.8 10.2^2

v = 23.5494

u = 123.509

b)
s = ?
u = 123.509
v = 0
a = -9.8
t = ?

0^2 = 123.509^2 + 2 -9.8 s

s= 778.289

c) reset 0 height to 750
s = ?
u = 23.5494 (v from part (a)
v = ?
a = -9.8
t = 1.5

s = 23.5494 1.5 - 1/2 9.8 1.5^2

s = 24.2991

>>5583478
it goes through as after 1.5 seconds it is 774.2991 high and max height is 778.289

as per >>5583521

>>5583521
Thanks, I get it now.

>>c) Calculate the thickness of the cloud layer

This isn't solvable. You either need to reword that to ask

>c) Calculate the *minimum* thickness of the cloud layer

OR

Assume that the probe hits the top of the cloud layer. In which case

> C) = B) - 750m