Projectile Motion (1 Viewer)

[lyounamu]

http://www.boredofstudies.org/courses/maths/3u/1095099280_2004_Mathematics_Extension_1_Trial_Paper_CSSA.pdf

PAGE 6 last question (Q 7) of the CSSA paper.

I don't understand the part ii and iii.

Help will be appreciated. Thanks in advance.

By the way, can anyone clarfy this for me?

Page 6 Q 6 b) ii)

I don't know what it means by the possible values of x . Are these just random numbers? ? (as long as they satisfy the equation I found in part i)

 

[eskimoh]

in 6b) its just refering to the range which the particle oscilates between. ie its two extremities
and yeh farrout i tried doing 7 pt ii) a few days ago! and i couldnt get it. i tried subbing in t=1 and assuming that they collide then and i couldnt get anywhere.. and then i tried finding the point at which they collide but i dont have any additional info for where they land or anything (to find theta or something) so i cant do it!!
sorry

let me know if you do find a solution thoo

 

[lyounamu]

in 6b) its just refering to the range which the particle oscilates between. ie its two extremities
and yeh farrout i tried doing 7 pt ii) a few days ago! and i couldnt get it. i tried subbing in t=1 and assuming that they collide then and i couldnt get anywhere.. and then i tried finding the point at which they collide but i dont have any additional info for where they land or anything (to find theta or something) so i cant do it!!
sorry

let me know if you do find a solution thoo

For 6b) ii) I got everything in Q 6. I just don't understand what they are talking about as in "the set of possible values of x". So you think it means extremeties? It doesn't say anything though.

I got q7.

I will post up soon.

Ok, first you find all the horizontal and vertical velocities for both particles.

Now, you compare both horizontal displacements.

x = 10 square root of (5) . cos@ t
And x = 10t
Now equate, 10t = 10 square root of (5) . cos@ t
10 = 10 square root of (5) cos@
cos@ = 10/(10square root of (5))
So @ = 63.4349488...

Then you do the same thing for both vertical displacements.
y = -1/2 g . t^2
and y = 10 square root of (5) sin@ . t - 1/2 g. t^2
Equate both velocities so you get,

10 square root of (5) sin@ . t - 20 = 0
Now substitute the angle we found, then 20t - 20 = 0
so t = 1.

I don't know how I worked it out. Once I read your post, I felt the surge of ideas in my head so I ended up working this out in 5 minutes. lol

Thanks for your post.

Now, substitute that to the vertical velocity which is y = 10square root of 5 . sin@ . t = 20