HSC Physics Question Thread

[nancy_cc]

Legend Nancy! Thanks for the help!

Hahaha no problem!!

[Yasminpotts1105]

"Draw all forces acting on the mass of a pendulum, when its velocity is zero, but within a swing."

Are the only forces I need gravity and tension in the string?

[jakesilove]

"Draw all forces acting on the mass of a pendulum, when its velocity is zero, but within a swing."

Are the only forces I need gravity and tension in the string?

Yep! Those are the only two relevant forces

[Yasminpotts1105]

"Discuss safety requirements for astronauts in regard to low gravitational forces in space."

So far I just have information about blackouts and red-outs and understand about the g-forces but I am not sure what else I need to say.

[jakesilove]

"Discuss safety requirements for astronauts in regard to low gravitational forces in space."

So far I just have information about blackouts and red-outs and understand about the g-forces but I am not sure what else I need to say.

That's probably plenty! If you want to discuss Astronauts whilst in space, think about the need for them to exercise so as not to lose too much muscle mass. But honestly, it sounds like you've got the gist of the dotpoint!

[teapancakes08]

Having trouble with this question:

An ‘extreme’ cyclist wants to perform a stunt in which he rides up a ramp, launching himself into the air, then flies through a hoop and lands on another ramp. The angle of each ramp is 30.0° and the cyclist is able to reach the launch height of 1.50 m with a launching speed of 30.0 km/h. Calculate:
(a) the maximum height above the ground that the lower edge of the hoop could be placed
(b) how far away the landing ramp should be placed.

I think I can work out the distance okay, but the finding of the max. height concerning the lowest edge of the hoop throws me off a bit. I might be overthinking this though. Does the question just require the usual equations or is there more math to it? If anyone can help me out, thank you so much!

[jamonwindeyer]

Having trouble with this question:

An ‘extreme’ cyclist wants to perform a stunt in which he rides up a ramp, launching himself into the air, then flies through a hoop and lands on another ramp. The angle of each ramp is 30.0° and the cyclist is able to reach the launch height of 1.50 m with a launching speed of 30.0 km/h. Calculate:
(a) the maximum height above the ground that the lower edge of the hoop could be placed
(b) how far away the landing ramp should be placed.

I think I can work out the distance okay, but the finding of the max. height concerning the lowest edge of the hoop throws me off a bit. I might be overthinking this though. Does the question just require the usual equations or is there more math to it? If anyone can help me out, thank you so much!

Hey there! I'm having a bit of trouble picturing this; do you interpret this as the ramp being 1.5 metres high at the point where the cyclist leaves the ramp? That's how I interpret it, let me know if you think the same.

In that case, just the normal math!! I'm super happy to show you the steps, but just in case you wanted to have a go yourself first, it would just be:

1- Resolve the velocity into horizontal and vertical components
2- Use \(v=u+at\) to find when it reaches the maximum height
3- Use \(\Delta y=ut-\frac{1}{2}at^2\) to find the maximum height itself

Then remember to add 1.5 metres to that number, because we started that high off the ground to begin with

If you are having trouble let me know and I'll show you the steps!

[teapancakes08]

Hey there! I'm having a bit of trouble picturing this; do you interpret this as the ramp being 1.5 metres high at the point where the cyclist leaves the ramp? That's how I interpret it, let me know if you think the same.

In that case, just the normal math!! I'm super happy to show you the steps, but just in case you wanted to have a go yourself first, it would just be:

1- Resolve the velocity into horizontal and vertical components
2- Use \(v=u+at\) to find when it reaches the maximum height
3- Use \(\Delta y=ut-\frac{1}{2}at^2\) to find the maximum height itself

Then remember to add 1.5 metres to that number, because we started that high off the ground to begin with

If you are having trouble let me know and I'll show you the steps!

Seems to be the case. The answers I got were (a) 2.39m (2.d.p) and (b) 6.14 (2.d.p), which also happen to be the answers when checking the back of the textbook. Thanks for helping me out 

[jamonwindeyer]

Seems to be the case. The answers I got were (a) 2.39m (2.d.p) and (b) 6.14 (2.d.p), which also happen to be the answers when checking the back of the textbook. Thanks for helping me out 

Cool beans!! Not a worry at all

[katnisschung]

stuck again 
2) A projectile is fired from the top of a 120m high cliff at 25m/s
it lands on the ground 6.4 s after firing

find
a) the initial velocity of the projectile
b) its initial horizontal velocity
c) its initial vertical velocity
d) its range
e) its maximum height
f) the time it takes to reach max height
g) its time of flight

thanks!

[jakesilove]

stuck again 
2) A projectile is fired from the top of a 120m high cliff at 25m/s
it lands on the ground 6.4 s after firing

find
a) the initial velocity of the projectile
b) its initial horizontal velocity
c) its initial vertical velocity
d) its range
e) its maximum height
f) the time it takes to reach max height
g) its time of flight

thanks!

Hey! This is a behemoth of a question; I've attached my solutions, but I totally understand if anything isn't clear, so if you need clarification just ask!

[katnisschung]

thank you!!!!!!!!!!!!!!

[katnisschung]

i can see that the answers are correct
but why is change in y considered -120m?
because i keep on picturing the distance from
the peak of the projectile to the bottom
so wouldn't it be >120?

[jakesilove]

i can see that the answers are correct
but why is change in y considered -120m?
because i keep on picturing the distance from
the peak of the projectile to the bottom
so wouldn't it be >120?

Basically, you just have to make sure you define a direction, and stick with it. I made down negative, and up positive. That meant that acceleration due to gravity was negative (as it pushes a particle DOWN), and that distance was negative if it traveled in the downwards direction!

[katnisschung]

hi jake
i understand that its negative
let me rephrase
wouldnt change in y be
<-120m becos the projectile travels above the building
then to the ground?

sorry if this isn't clear