Maths/science A Level question... help???

squarehead94

squarehead94

squarehead94

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hey guys, so i was reading through some old science papers today, and spotted this. any of you guys able to work it out? you are a better man than I if so.

A cube measuring 0.92 square metres, weighing 101.6kg is pushed from a plane at 1371.6 metres.

1) What is the time of the freefall ??
2)What will be the speed just before impact ??
3)Will the cube reach terminal velocity, if yes how long after the start ??

cheers if you can help!!
 
1) SUVAT - you know the distance, starting velocity and acceleration
2) knowing the time taken you can use another SUVAT equation to calculate the speed (again you know the acceleration and initial velocity)
3) For terminal velocity, the acceleration = 0, the forces must balance, mg=air resistance upward. Are you not given any more info?
 
nope - this is all the info in the question i got. although, you have provided more information than i dare dreamed have producing. i stick to my strong subject, business :P

so is there like, an answer to the question? like, a number?
 
without being given some equation for air resistance this question is impossible.

if you assume the cube is falling in a vacuum (ie, ignore air resistance) then you could use simple suvat equations but there are two problems with this. the main one is that terminal velocity would be the speed of light and the other problem is that no aeroplane can fly in a vacuum.

i've looked into it quickly to see if you can determine the air resistance with the information you have but, without knowing the material of the cube or an equation for air temperature vs height then this is impossible too

*edit*
do you have a formula book somewhere that gives simplified equations or constants for air resistance?
 
What paper is it? Some exam papers had essential equations and revelant information at the front?
 
s = displacement
u = initial velocity
v = velocity
a = acceleration
t = time.



you know a few of them initially.
s = 1371.6
u = 0
a = 9.8ms^-1 assuming downwards is positive. (mavity, change 9.8 to 9.81 if higher precision is required)
and you want to find t.

s = ut + 1/2(a)(t)^2.

u = 0 so first part cancels.

s = 1/2 at^2

2s = at^2

2s/a = t^2

SQRT(2s/a) = t

SQRT(2x1371.6/9.8) = t


16.730761110808401459793199112639

or 16.7 to 3sf.

I haven't done that sort of maths in a while so I might have made mistakes or what not.


now you have s u a and t and need to find v, the final velocity.

so in the exam you don't want to use the value you worked out (t) in case you got it wrong. so use S U and A again.

another suvat equation is

v^2 = u^2 + 2as

v^2 = 0^2 + 2x9.8x1371.6

v^2 =26883.36

v = SQRT(26883.36)

v = 163.96145888592233430597335130386

v = 164 ms^-1


as for the air resistance question, as said above it's hard to do without knowing more information.
 
Mason- said:
s = displacement
u = initial velocity
v = velocity
a = acceleration
t = time.



you know a few of them initially.
s = 1371.6
u = 0
a = 9.8ms^-1 assuming downwards is positive.
and you want to find t.

s = ut + 1/2(a)(t)^2.

u = 0 so first part cancels.

s = 1/2 at^2

2s = at^2

2s/a = t^2

SQRT(2s/a) = t

SQRT(2x1371.6/9.8) = t


16.730761110808401459793199112639

or 16.7 to 3sf.

I haven't done that sort of maths in a while so I might have made mistakes or what not.


now you have s u a and t and need to find v, the final velocity.

so in the exam you don't want to use the value you worked out (v) in case you got it wrong. so use S U and A again.

another suvat equation is

v^2 = u^2 + 2as

v^2 = 0^2 + 2x9.8x1371.6

v^2 =26883.36

v = SQRT(26883.36)

v = 163.96145888592233430597335130386

v = 164 ms^-1
Click to expand...

Blimey. I'm not clever enough to know if this is correct or not.
 
[FnG]magnolia;22279448 said:
Blimey. I'm not clever enough to know if this is correct or not.
Click to expand...

I thought you were quite educated Magnolia, perhaps not in maths.

They are standard equations of motion. Taught in GCSE physics and A-Level mathematics.
 
[FnG]magnolia;22279465 said:
I sat my O grades/GCSEs in the eighties. It's funny the stuff you forget or, as I suspect in this case, never knew.
Click to expand...

I've done nothing maths related in a year now (just done my first year of uni) and I've forgotten so much of my maths from A-Level which sucks because I wish I took Maths at university now.
 
Mason looks fine to me at first glance, it might be worth here stating all the kinematic equations so the OP knows which ones he could use:

v = u + at
v^2 = u^2 + 2as
s = ut + 1/2at^2

Obviously rearranging the equations to get variables you need.
 
Refering to an equation I've just found in a book, for terminal velocity, use the drag equation

F= 1/2 * density * V^2 * drag coefficient * area (in meters squared)

Take the drag coefficient as about 1.05 or 1, then rearange to find v, and sub F in as the acceleration due to mavity, from this you'll find terminal velocity. Hope this helps. (You can look up the density of air) But its 1kg/m^3 IIRC
 
[FnG]magnolia;22279448 said:
Blimey. I'm not clever enough to know if this is correct or not.
Click to expand...

It's not that hard, it just looks hard.

The hardest bit is remembering the equation, which I suspect he might of looked up.

Then it's just some rearranging the formula gcse style...
 
Last edited:
Mason- said:
s = displacement
u = initial velocity
v = velocity
a = acceleration
t = time.
Click to expand...

Nope, you can't use those formulae as they assume constant acceleration. Drag is implied to be non-negligible so acceleration varies with velocity squared.

Start with Newton.

F = ma

kv^2 - mg = ma

(k/m).v^2 - g = dv/dt

Solve this lovely differential equation and you'll get velocity as a function of time. You'll also need a value for k.
 
im asking now for a value for the air resistance. i dont study these subjects, I just found some old papers and took an interest. thanks for the help so far guys!
 
Morbius said:
Nope, you can't use those formulae as they assume constant acceleration. Drag is implied to be non-negligible so acceleration varies with velocity squared.

Start with Newton.

F = ma

kv^2 - mg = ma

(k/m).v^2 - g = dv/dt

Solve this lovely differential equation and you'll get velocity as a function of time. You'll also need a value for k.
Click to expand...

I was thinking that after I posted :) acceleration cant just be mavity
 
F = -1/2 C rho A v V
Fd -> the drag force in vector form [Fdx; Fdy]
C -> the drag coefficient (0.3 for a baseball)
rho-> the density of air (1.3 kg / m^3)
A -> cross sectional area of the object (0.0042 m^2 for a baseball)
v -> the magnitude of the current velocity
V -> the current velocity in vector form [Vx, Vy]

( I AM REALLY UNSURE IF YOU CAN USE THIS AS I HAVE NEVER STUDIED THIS).
Click to expand...
That might be useful, you an get the drag coefficient as it's a square, you know the cross sectional area then you equate it to mg.
mg=1/2CrhoAV^2
Take the RHS under
2mg/CrhoA=V^2
sqrt(2mg/CrhoA)=V
Then using suvat you can work out S
V^2=U^2+2AS
V^2=2AS
V^2/2A=S
 
RomanNose said:
F = -1/2 C rho A v V

That might be useful, you an get the drag coefficient as it's a square, you know the cross sectional area then you equate it to mg.
mg=1/2CrhoAV^2
Take the RHS under
2mg/CrhoA=V^2
sqrt(2mg/CrhoA)=V
Then using suvat you can work out S
V^2=U^2+2AS
V^2=2AS
V^2/2A=S
Click to expand...

No, because v^2=u^2+2as assumes constant acceleration. Your acceleration isn't. :)
 
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