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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: t5am94 on May 06, 2011, 11:26:12 pm

Title: NEED HELP WITH THIS QUESTION!
Post by: t5am94 on May 06, 2011, 11:26:12 pm

I couldn't think of a way to work this out. Maybe one of your guys out there can help me out.

A helicopter is moving vertically upwards at 5ms. When it is 150m above the ground a package is released.

(a) What is the ACCELARATION of the package when it reaches maximum height?
(b) How long does it take to reach maximum height?

Cheers.

Title: Re: NEED HELP WITH THIS QUESTION!
Post by: schnappy on May 06, 2011, 11:37:27 pm

Acceleration is just gravity, so a = -9.8 where the negative direction is towards Earth's centre of mass.

For max height, v=o u=5 a=-9.8 t=?
Use the kinematics equations to find t. v = u + at

Title: Re: NEED HELP WITH THIS QUESTION!
Post by: onur369 on May 07, 2011, 12:30:15 am

for max height you can also do, hmax=u^2/2g  so 25/20 which is 1.25

Title: Re: NEED HELP WITH THIS QUESTION!
Post by: VivaTequila on May 10, 2011, 10:26:03 pm

Acceleration is always a constant 9.8m s-1 down (to the earth). When you throw a ball up and down, when it's at max height, it's acceleration is 9.8m s-1 down as well, so why is it any different for a package?

For the time it takes to reach maximum height, you know that it is decelerating at 9.8ms-1 in the absense of any other forces and has an initial velocity of 5m s-1.
When it reaches the top point, it has stopped moving, so you need to find how long it takes for the velocity to hit zero.

V=U+AT
(0)=(5)+-9.8*T
5=9.8T
T=5/9.8
T=0.51020408163265306122448979591837s (just copied out of comp calc)


can anyone confirm this?

Title: Re: NEED HELP WITH THIS QUESTION!
Post by: Thu Thu Train on May 10, 2011, 11:16:55 pm

Quote from: VivaTequila on May 10, 2011, 10:26:03 pm

Acceleration is always a constant 9.8m s-1 down (to the earth). When you throw a ball up and down, when it's at max height, it's acceleration is 9.8m s-1 down as well, so why is it any different for a package?

For the time it takes to reach maximum height, you know that it is decelerating at 9.8ms-1 in the absense of any other forces and has an initial velocity of 5m s-1.
When it reaches the top point, it has stopped moving, so you need to find how long it takes for the velocity to hit zero.

V=U+AT
(0)=(5)+-9.8*T
5=9.8T
T=5/9.8
T=0.51020408163265306122448979591837s (just copied out of comp calc)


can anyone confirm this?

This is right but you'd normally only have to go to 0.51s :P

Title: Re: NEED HELP WITH THIS QUESTION!
Post by: onur369 on May 10, 2011, 11:25:54 pm

(a)What is the ACCELARATION of the package when it reaches maximum height?
(b) How long does it take to reach maximum height?

a) v = (u^2+2gh)^.5   (25+2x10x150)^.5 = 55
 v^2=u^2+2ax
3025=25+2xAx150
a=10ms^2

b) tup= u/g = 5/10 = 0.5 seconds.

I think they are both correct.

Title: Re: NEED HELP WITH THIS QUESTION!
Post by: schnappy on May 10, 2011, 11:28:20 pm

For part a, how is it anything other than -9.8 ms^-2?

Title: Re: NEED HELP WITH THIS QUESTION!
Post by: onur369 on May 11, 2011, 05:28:02 pm

check my solution, i miss calculated it before.