Answer:
a) 
b) 
Explanation:
Given:
- upward acceleration of the helicopter,

- time after the takeoff after which the engine is shut off,

a)
<u>Maximum height reached by the helicopter:</u>
using the equation of motion,

where:
u = initial velocity of the helicopter = 0 (took-off from ground)
t = time of observation


b)
- time after which Austin Powers deploys parachute(time of free fall),

- acceleration after deploying the parachute,

<u>height fallen freely by Austin:</u>

where:
initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)
time of free fall


<u>Velocity just before opening the parachute:</u>



<u>Time taken by the helicopter to fall:</u>

where:
initial velocity of the helicopter just before it begins falling freely = 0
time taken by the helicopter to fall on ground
height from where it falls = 250 m
now,


From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.
<u>remaining time,</u>



<u>Now the height fallen in the remaining time using parachute:</u>



<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>



Yes you can, with using scientific experiment.
Ask a question -- Do background Research -- Construct a Hypothesis --Test with an Experiment -- Procedure working? -- Yes or no? -- Analyze Data and Draw Conclusions
With an experiment you can discover if its correct or not.
Hope this helps ! <3
The acceleration rate would be .14667 m/s^2
I think it would be WEIGHT and ROUGHNESS OF SURFACE.