Problem help (1 Viewer)

[theBOD]

A balloon has been ascending vertically from the ground at a constant speed for 10.0 s when a stone is allowed to fall from the side of the balloon. If it takes the stone a further 5.00 s to reach the ground, find

a) the speed of the balloon; and
b) its height at the instant that the stone is released

ANS: a) 8.17 ms^-1 b) 81.7m

Help me work this out please

 

[someth1ng]

I don't think there is enough information.

 

[romesh]

There is enough information I'm not 100% sure about how you would go about solving it using HSC physics equations. The mathematics way of doing this problem is:

Suppose that the stone is dropped at time t=0. Let the balloon velocity be v_0. The stone starts out at a height x=10*v_0 (the velocity multiplied by the time it was ascending for). Gravitational acceleration reduces the balloon's velocity by 9.8m/s every second. So, we can write

v(t) = v_0 - 9.8*t <--- This equation provides the velocity of the stone at time t

Now, integrate the velocity to find the position as a function of t

x(t) = v_0*t - 9.8/2*t^2 + c

We said the balloon starts out at t=0, x = 10*v_0 so if we substitute t=0 then it is clear that c = 10*v_0. Thus

x(t) = v_0*t - 9.8/2*t^2 + 10*v_0

We are told that 5 seconds later, the stone hits the ground. Thus at t=5, x=0. Substituting these values in gives

0 = 5*v_0 - 9.8/2*(5^2) + 10*v_0

Rearranging gives

v_0 = 9.8*25/2/15 = 8.17 m/s

And since the balloon's original height was 10*v_0, the height was 81.7m

 

[Fizzy_Cyst]

You could do it using HSC Physics.

Just by doing the following:

delta_Y=Uy*t + 1/2ayt^2

From the question we are told that:

delta_Y = -10Uy (the stone 'falls' the displacement of the balloon, hence is negative)

Therefore:

-10Uy = Uy*t + 1/2ayt^2

The question says t = 5

-10Uy = 5Uy - 122.5

simplified to:

-15Uy = -122.5

Therefore Uy = -122/-15 = 8.17ms^-1

delta_Y = 8.17*10 = 81.7m

I just noticed this is pretty much the same as Romesh above, lolz.

Thx Romesh

 

[someth1ng]

Doesn't look like HSC Physics to me?
Actually, I re-read the question and didn't see the "from the ground" part so I thought it wasn't but it actually is.

My apologies.