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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: dusty_girl1144 on July 28, 2008, 09:32:35 pm

Title: The Hells Of Kinematics
Post by: dusty_girl1144 on July 28, 2008, 09:32:35 pm: mmmm ive tried for HOURS on end to do displacement, velocity and acceleration.... and y i dont get it i dunno. its diffing and intergrating. so heres the question...

from a balloon ascending with a velocity of 10 m/s a stone was dropped and reached the ground in 12 seconds. given that the gravitational acceleration was 9.8 m/s^2

find a. the height of the balloon when the stone was dropped
b. the greatest height the stone reached.

enjoy :) coz i sure as hell didnt....

answers:
a. 585.6m
b. 590.70m
Title: Re: The Hells Of Kinematics
Post by: dcc on July 28, 2008, 09:37:25 pm: Part A:

We know $a = 9.8 \mbox{m/s$^{-2}$},\ u = -10 \mbox{m/s},\ t = 12 \mbox{s}$ (considering downwards as positive)

Since $h = ut + \frac{1}{2}at^2 \implies h = -10 \cdot 12 + \frac{1}{2} \cdot 9.8 \cdot 144$

$h = 585.6$ i.e. The stone had to fall 585.6 metres to hit the ground.
Title: Re: The Hells Of Kinematics
Post by: dusty_girl1144 on July 28, 2008, 09:42:54 pm: Quote from: dcc on July 28, 2008, 09:37:25 pm
We know $a = -9.8 \mbox{m/s$^{-2}$},\ u = 10 \mbox{m/s},\ t = 12 \mbox{s}$

Since $h = ut + \frac{1}{2}at^2 \implies h = 10 \cdot 12 + \frac{1}{2} \cdot -9.8 \cdot 144$

$h = -585.6$ i.e. The stone had to fall 585.6 metres to hit the ground.

ok well that formular is to do with constant acceleration LOL. which is the NEXT exercise. but hey u got the answer so im happy :) thanxs heaps!

.xxx.
Title: Re: The Hells Of Kinematics
Post by: dusty_girl1144 on July 28, 2008, 09:47:38 pm: Part B on for sale now
Title: Re: The Hells Of Kinematics
Post by: dcc on July 28, 2008, 09:48:46 pm: Consider the following then:

For the stone:

$a = \dfrac{dv}{dt} = 9.8 \mbox{m/s$^{-2}$} \implies v = 9.8t + C$

Now at $t = 0$ , we know that $v = -10 \mbox{m/s}$ , hence we have:

$v = \dfrac{dx}{dt} = 9.8t - 10$

Now to find the height, we find the displacement in the first 12 seconds:

$h = \int_0^{12} \left(9.8t - 10\right)\ dt = \left[4.9t^2 - 10t\right]_0^{12} = 585.6 \mbox{m}$
Title: Re: The Hells Of Kinematics
Post by: dusty_girl1144 on July 28, 2008, 09:53:15 pm: seems so easy now.... :( but thanxs for showing me both ways :)
Title: Re: The Hells Of Kinematics
Post by: dcc on July 28, 2008, 09:55:18 pm: To find the greatest height, consider the function:

$x(t) = 10t - 4.9t^2 + 585.6$ where positive values represent upwards.

The maximum of x(t) occurs when $x'(t) = 0$

i.e. $0 = 10 - 9.8t \implies t = \frac{10}{9.8} \mbox{s}$

$x\left(\frac{10}{9.8}\right) = 590.7 \mbox{m}$

I apologise for changing the signs in this part of the question, this would not be an appropriate thing to do during a real exam, but unfortunately my brain changed it half way through :P
Title: Re: The Hells Of Kinematics
Post by: dusty_girl1144 on July 28, 2008, 10:09:30 pm: Quote from: dcc on July 28, 2008, 09:55:18 pm
To find the greatest height, consider the function:

$x(t) = 10t - 4.9t^2 + 585.6$ where positive values represent upwards.

The maximum of x(t) occurs when $x'(t) = 0$

i.e. $0 = 10 - 9.8t \implies t = \frac{10}{9.8} \mbox{s}$

$x\left(\frac{10}{9.8}\right) = 590.7 \mbox{m}$

I apologise for changing the signs in this part of the question, this would not be an appropriate thing to do during a real exam, but unfortunately my brain changed it half way through :P

where exactly do u sub the (10/9.8)?
Title: Re: The Hells Of Kinematics
Post by: dcc on July 28, 2008, 10:11:39 pm: Into $x(t) = 10t - 4.9t^2 + 585.6$ , as this is the function which gives the height of the stone above the ground as a function of time.
Title: Re: The Hells Of Kinematics
Post by: dusty_girl1144 on July 28, 2008, 10:15:58 pm: thanxs for helping me dcc :)