Baulkham Hills High Trial 2005 Projectile Motion Question (1 Viewer)

[Sloppy Sailor]

I'm having trouble with this question, mainly part b). Can you guys answer it?

<click the link to get a .jpg of the question>

http://img530.imageshack.us/img530/4207/physicsquestionwg8.jpg

Much appreciated -

~Sloppy~

 

[Forbidden.]

I don't have pen and paper here but I can tell you how to try b)

b)

Use v_y² = u_y² + 2 a_y Δy

Find Δy from the height of the tree i.e. 12 metres
a_y is ALWAYS a constant -9.8ms^-2 on Earth.
With regards to v_y², a projectile has no vertical motion at the vertex of a parabolic path.

Now make u_y the subject.

 

[jason21]

I use a basic equation to calculate this which you may not because it is not given on the formula sheet.

I use height (H)=V^2 (sin0)^2/2g

therefore, 12=V^2 (sin 30)^2/19.6

rearrange this so that V is subject, therefore V=sqrt 2Hg/(sin0)^2

This will give you an answer of approximately 30.67ms-1

 

[wrxsti]

I use a basic equation to calculate this which you may not because it is not given on the formula sheet.

I use height (H)=V^2 (sin0)^2/2g

therefore, 12=V^2 (sin 30)^2/19.6

rearrange this so that V is subject, therefore V=sqrt 2Hg/(sin0)^2

This will give you an answer of approximately 30.67ms-1

dont you just love 3unit math

 

[jason21]

no actually i hate it hence the general maths that i do lol. but i remember those equations cause they will help me alot

 

[lionking1191]

where's the 3u maths in that..?

 

[Forbidden.]

where's the 3u maths in that..?

y = - gt²/2 + Vtsinθ

y' = - gt + Vsinθ
Greatest height is reached when y' = v_y = 0
0 = - gt + Vsinθ
gt = Vsinθ
t = Vsinθ/g


Let greatest height y be h.
h = - gt²/2 + Vtsinθ
(t = Vsinθ/g)
h = [- g(Vsinθ/g)²/2] + V(Vsinθ / g)sinθ
h = - V²sin²θ/2g + V²sin²θ/g

Therefore the greatest height is given by:
h = V²sin²θ/g

 

[lionking1191]

isn't that 2u? i might be wrong, don't quote me