'06 HSC Projectile Question (1 Viewer)

[weirdguy99]

Could someone help with this question from the 2006 HSC, question 16. (I reworded this as it needed a diagram to explain the question)

A projectile leaves the ground at a horizontal velocity of 45m/s and a vertical velocity of 40m/s.

a)State the horizontal component of the projectile’s velocity when it lands.
b)Find the magnitude of the initial velocity of the projectile.
c)Calculate the maximum height attained by the projectile.
d)Calculate the range of the projectile, if it lands level with its starting position.

I got the answers, just need someone to help confirm them =)

 

[Pwnage101]

a) same as initial: 45m/s( to the right)
b)sqrt(45^2+40^2)=60.20797289...=60m/s (2 sig fig)
c)v=u+at ---> t=0-40/-9.8=4.081632, Sy=ut+0.5at^2=81.63265...=82m (2 sig fig)
d)time of flight = 2*t=8.163265... Sx=ut=367.3469388...=370m (2 sig fig)

 

[cutemouse]

Could someone help with this question from the 2006 HSC, question 16. (I reworded this as it needed a diagram to explain the question)

A projectile leaves the ground at a horizontal velocity of 45m/s and a vertical velocity of 40m/s.

a)State the horizontal component of the projectile’s velocity when it lands.
b)Find the magnitude of the initial velocity of the projectile.
c)Calculate the maximum height attained by the projectile.
d)Calculate the range of the projectile, if it lands level with its starting position.

I got the answers, just need someone to help confirm them =)

a) 45 m/s
b) m/s

c) v=u+at

At max height v=0
So 0=40 - 9.8t
t= 4.08 s

Now


d)

Probs did some stupid error somewhere, meh...

 

[weirdguy99]

Yep, got it right. Thanks for the input guys.

 

[weirdguy99]

Another one if anyone is there.. from `06 HSC, question 18

An object is stationary in space and located at a distance 10 000 km from the centre of a certain planet. It is found that 1.0 MJ of work needs to be done to move the object to a stationary point 20 000 km from the centre of the planet.

Calculate how much more work needs to be done to move the object to a stationary
point 80 000 km from the centre of the planet.

 

[hursalas05]

Sorry but would the easier way to go about finding the maximum height be by using:

v^2 = u^2 + 2ar

being

0 = (40)^2 + 2 (-9.8) r

therefore straight away giving you the value of r being the distance?

I know that the answer will be the same answer, but what do you guys think would be the quicker way to get the solution?

 

[Pwnage101]

Sorry but would the easier way to go about finding the maximum height be by using:

v^2 = u^2 + 2ar

being

0 = (40)^2 + 2 (-9.8) r

therefore straight away giving you the value of r being the distance?

I know that the answer will be the same answer, but what do you guys think would be the quicker way to get the solution?

Definitely.

 

[Dragonmaster262]

Another one if anyone is there.. from `06 HSC, question 18

An object is stationary in space and located at a distance 10 000 km from the centre of a certain planet. It is found that 1.0 MJ of work needs to be done to move the object to a stationary point 20 000 km from the centre of the planet.

Calculate how much more work needs to be done to move the object to a stationary
point 80 000 km from the centre of the planet.

Tricky question

Work done is the change in potential energy.

1 MJ = 1x10^6 J

Chenge in EP = (-Gm1m2)/(20000x10^3) - (-Gm1m2)/(10000x10^3)
1x10^6 = (-Gm1m2+Gm1m2)/(20000x10^3)
2x10^13 = Gm1m2

To move object from 20000 km to 80000 km

-Gm1m2/(80000x10^3) - (-Gm1m2)/(20000x10^3) (Now sub Gm1m2 value in)

Change in EP = -250000 + 1000000 = 750000 J

Am I right?

 

[weirdguy99]

I think you read the question wrong. It says to move the object 80000k from the centre of the planet, so its not 20000km to 80000km, its 0km to 80000km.

 

[Dragonmaster262]

Damn but my method is right isn't it?