wildareal's questions thread

[wildareal]

Hey thanks. For part c) ii) I know you have to make c in terms of A and t, but then what do you do next?

[wildareal]

A hot air balloon is ascending at a speed of 2m/s. When the hot air balloon is at a height above the ground of 300 metres, a passenger accidentally drops their wallet over the side of the balloon basket.
(a)   Assuming no air resistance, how long does it take the wallet to strike the ground? Give your answer correct to 2 decimal places.

[pi]

A hot air balloon is ascending at a speed of 2m/s. When the hot air balloon is at a height above the ground of 300 metres, a passenger accidentally drops their wallet over the side of the balloon basket.
(a)   Assuming no air resistance, how long does it take the wallet to strike the ground? Give your answer correct to 2 decimal places.

Don't remember the formulas, but this should help:

(down = +ve)

a = 10, u = -2, x = 300, t = ?

[xZero]

x=ut+0.5at^2, define going downward is positive, x=300,u=-2,a=9.81, sub in the numbers and solve for t.

spesh uses a=9.81 right?

[b^3]

Yeh 9.8m/2^s is right, well it is what essential use anyway.

[pi]

spesh uses a=9.81 right?

*facepalm*

curse you physics!

yeh, its defs 9.8 ms^-2

[wildareal]

In an effort to save the wallet the burners on the balloon are turned off. The balloon stops rising and remains steady at an altitude of 350 metres. The wallet’s owner dons an emergency parachute and goes overboard. Because of the low altitude he releases the parachute immediately.
(b)   (i) Assuming that the retardation due to air resistance is equal to 1.2v2, where v m/s is the   velocity at time t seconds, express the acceleration of the parachutist in terms of his velocity.
(ii) Hence express the velocity in terms of the position x metres below the parachutist’s starting point and find the speed of the parachutist as he hits the ground

As defined by Rohit: up is neg, down is pos: dv/dt=9.8-1.2v^2

How do you do ii)?

Thanks.

[xZero]

dv/dt = dv/dx * dx/dt -> dv/dt = dv/dx * v -> dv/dt=v*dv/dx, sub this into your equation, rearrange, integrate it with respect to v then rearrange it back to make v the subject

[tony3272]

In an effort to save the wallet the burners on the balloon are turned off. The balloon stops rising and remains steady at an altitude of 350 metres. The wallet’s owner dons an emergency parachute and goes overboard. Because of the low altitude he releases the parachute immediately.
(b)   (i) Assuming that the retardation due to air resistance is equal to 1.2v2, where v m/s is the   velocity at time t seconds, express the acceleration of the parachutist in terms of his velocity.
(ii) Hence express the velocity in terms of the position x metres below the parachutist’s starting point and find the speed of the parachutist as he hits the ground

As defined by Rohit: up is neg, down is pos: dv/dt=9.8-1.2v^2

How do you do ii)?

Thanks.

Beaten.









Whoops i forgot the 1.2 in front of the . 

[wildareal]

(cosec^2(x)-1)*(cos(TT/2-x)*tan(x)

[b^3]

If you want it simplified (unless I missed a post) then you need to know this and
so this will give you
then or equivalent

[tony3272]

If you want it simplified (unless I missed a post) then you need to know this and
so this will give you
then or equivalent

This is the right method but wrong identity.
The identity is:


so you get:

[b^3]

Whoops sorry my mistake, I always use the oppostie one for some reason.

[luken93]

Whoops sorry my mistake, I always use the oppostie one for some reason.

Just have cos^2(x) + sin^2(x) = 1
Divide them by sin^2 to get cot^2 + 1 = cosec^2
Divide by cos^2 to get 1 + tan = sec

[b^3]

Whoops sorry my mistake, I always use the oppostie one for some reason.

Just have cos^2(x) + sin^2(x) = 1
Divide them by sin^2 to get cot^2 + 1 = cosec^2
Divide by cos^2 to get 1 + tan = sec

Yeh I know, I have a habit of writing tan(x) for cos(x)/sin(x) instead of cot(x)