STAV 2007 trial exam Questions

[cranberry]

just did my first timed practice exam! thought i aced it (got 67%  ) oh well practic..  :coolsmiley:



Question 6
A student celebrating the completion of his examinations takes
an early morning champagne and chicken balloon flight over
Melbourne. At one point in the trip the balloon is rising at a
steady speed of 4.0 m s-1 and is at an altitude of 150 m above the
ground. At this point the unlucky student drops the only bottle
opener in the balloon over the side! Fortunately the champagne
can still be opened.
How long does it take for the bottle opener to reach the ground?

Is 4m/s up, as a constant speed, zero when put into motion equations (as i did below)?

i did:
x=150
u=0
t=?
a=10

and got 5.5s. the answer used u=4 and got 5.9s...does it matter?

ive got a few more which the answers dont explain that well

exam/solutions attached

ty

[Thu Thu Train]

Yeah it matters. The initial velocity of the bottle opener was 4m/s up its not the same as saying the initial velocity is zero.

[Vincezor]

taking 'up' as postive





use quadratic formula



But yes, you must use u=4m/s as the balloon is rising at that speed and the student is in that balloon... so the student is rising at as well

[cranberry]

okay ty.

also this one:

Janette is competing at an athletics track and field meeting
in the high jump.
Janette, who weighs a petite 45 kg, manages to jump a
height of 1.9 metres.
Janette runs towards the bar and pushes off from the ground
at an angle of 80° to the horizontal.
Question 7
What is Janette’s take off speed? (Treat Janette as having projectile motion)

Question 8
How far horizontally has Janette travelled from her take off point when she lands? You can
assume that she lands at the same vertical height that she takes off from.


for Q7 i did:

x=1.9m
a=10
u=?
v=0

using v^2=u^2 + 2ax....idk why they use u

= U sin 80°

i got 6.16m/s (im getting answers just off theirs so idk if my way is right aswell..)

ty

[Vincezor]

7. Usin80 is just the vertical component of take off speed (Remember projectile motion?)

8.

Horizontally,



However as it is horizontal component, u = v, therefore



So you just find u(horizontal) = ucos80 and T(total) = 2t where t= time taken to reach the top (Time up = Time down)

and just multiply those together to get your answer?

[cranberry]

ah okay. i used the vertical component as the final answer instead of finding the hypotenuse>.<