2012 vcaa examination?

[CatMeoooow]

Question 6 
Fred throws a ball at an angle of 60(degrees) to the horizontal so that it rises to a vertical distance of 15m. The landing point is the same hight as the starting point. 
Calculate the speed of which the ball leaves freds hand?        Answer: 20m/s   

Tell me if I'm correct     
 
v=0     u=Usin60(degrees)  a=-10   x=15   

v^2=u^2+2ax   

then I rearranged to give    me     

Squareroot of 300/sin60^2   which gave me an answer of 20.     

The next question askes calculate the time of flight of the ball?   
Can someone help me with this? 

[Lasercookie]

To figure out the time of flight, should do. We can split the motion up in two, once going up, and once going down. The time to go up is the same time to drop back down to where it originally started.

When it reaches it's maximum height, , the velocity at that very point will be zero,

It's acceleration is of course g, so .

(note that the vt term dropped out, as v = 0)







For the total time of flight, multiply by 2,

Significant figures, hmmm, 2 significant figures. so you'd state your final answer as

edit: added decimal answers

[CatMeoooow]

Thanks that was very clear

[CatMeoooow]

Just curios do you multiply by 2 because its the same time going up and same going down(as you stated above), thus giving you the total flight?

[Lasercookie]

Just curios do you multiply by 2 because its the same time going up and same going down(as you stated above), thus giving you the total flight?

Yep.

We can take it the long way around too. So we already know that the time to go up is , let's calculate the time to fall back down. When it's falling back down, at the start it's velocity is zero. It's acceleration is g. The height it'll drop is 15m. Since it's dropping back down, it's displacement will be



(you might be able to see by now we'll get the same result)







So to get the total time of motion just add the going up and going down times.

[CatMeoooow]

I found it an even quicker, because the path is parabolic we could just use  2xuxsintheta/g       2x20xsin60/10   = 3.464101615

[Lasercookie]

I found it an even quicker, because the path is parabolic we could just use  2xuxsintheta/g       2x20xsin60/10   = 3.464101615

Nice work.

Yeah, those derived formulas. You can use them by all means on exams, but I (personally) felt it's always safer to not rely on them and do it the "regular" way which forces you (to an extent) to think a bit about the situation and possibly come to a realisation about a tricky point in the question that you may have overlooked at first.

What is especially important since those derived formulas only apply to specific situations, which a condition given in the question could easily make invalid - you need to ensure that you're applying the correct formulas to the correct situations.

And also the fact that there's probably not too many questions where they're actually valid, it's better to know inside out the "regular" method so you can tackle most questions VCAA will throw at you.

[CatMeoooow]

Just one more thing, It just confuses me a little bit on 2011's exam with the projectile motion question about the cannon how you can find the total time by using the equation x=ut+1/2at^2 while for this question you have to break it down into two, to find time. 

How do you know when to breakdown into two segments and times by 2 to find total time?

[Lasercookie]

Just one more thing, It just confuses me a little bit on 2011's exam with the projectile motion question about the cannon how you can find the total time by using the equation x=ut+1/2at^2 while for this question you have to break it down into two, to find time. 

How do you know when to breakdown into two segments and times by 2 to find total time?

It shouldn't confuse you that there are multiple methods to get the same answer. That alternate method you stated works for both questions. We know when to multiply by 2 depending on what approach we use to solve the question.

With those projectile motion questions, such as the cannon, we can split the vectors up into their vertical and horizontal components.

We can analyse the vertical motion for the cannon and finding the time it takes to get to the peak (using v = u + at for example) - and then doubling by 2 since in this case it's symmetrical (it'd take the same time to go down). Before for the ball question, we analysed the vertical component only too, we used x = ut+1/2at^2 though, since that's what was easiest with the values we were given.

You can do a similar method if you were to analyse the horizontal component too.

So, if you wanted to use for the cannon question, you could note that the overall vertical displacement is 0, and hence go:

and then solve that quadratic equation, reject the solution for t =0. That'd get you the right answer, but it's a bit more effort to do that by hand.

The same applies for the 2012 question, overall vertical displacement is zero. The vertical component of the initial  velocity is



Solve and  reject the solution for t =0, giving you

Personally, I don't want to deal with things like factorising quadratics and quadratic formula when there's a much easier method. For the 2011 question:



and multiply by 2 to get