Respuesta :
Answer:
a) [tex]h=250\ m[/tex]
b) [tex]\Delta h=0.0835\ m[/tex]
Explanation:
Given:
- upward acceleration of the helicopter, [tex]a=5\ m.s^{-2}[/tex]
- time after the takeoff after which the engine is shut off, [tex]t_a=10\ s[/tex]
a)
Maximum height reached by the helicopter:
using the equation of motion,
[tex]h=u.t+\frac{1}{2} a.t^2[/tex]
where:
u = initial velocity of the helicopter = 0 (took-off from ground)
t = time of observation
[tex]h=0+0.5\times 5\times 10^2[/tex]
[tex]h=250\ m[/tex]
b)
- time after which Austin Powers deploys parachute(time of free fall), [tex]t_f=7\ s[/tex]
- acceleration after deploying the parachute, [tex]a_p=2\ m.s^{-2}[/tex]
height fallen freely by Austin:
[tex]h_f=u.t_f+\frac{1}{2} g.t_f^2[/tex]
where:
[tex]u=[/tex] initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)
[tex]t_f=[/tex] time of free fall
[tex]h_f=0+0.5\times 9.8\times 7^2[/tex]
[tex]h_f=240.1\ m[/tex]
Velocity just before opening the parachute:
[tex]v_f=u+g.t_f[/tex]
[tex]v_f=0+9.8\times 7[/tex]
[tex]v_f=68.6\ m.s^{-1}[/tex]
Time taken by the helicopter to fall:
[tex]h=u.t_h+\frac{1}{2} g.t_h^2[/tex]
where:
[tex]u=[/tex] initial velocity of the helicopter just before it begins falling freely = 0
[tex]t_h=[/tex] time taken by the helicopter to fall on ground
[tex]h=[/tex] height from where it falls = 250 m
now,
[tex]250=0+0.5\times 9.8\times t_h^2[/tex]
[tex]t_h=7.1429\ s[/tex]
From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.
remaining time,
[tex]t'=t_h-t_f[/tex]
[tex]t'=7.1428-7[/tex]
[tex]t'=0.1428\ s[/tex]
Now the height fallen in the remaining time using parachute:
[tex]h'=v_f.t'+\frac{1}{2} a_p.t'^2[/tex]
[tex]h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2[/tex]
[tex]h'=9.8165\ m[/tex]
Now the height of Austin above the ground when the helicopter crashed on the ground:
[tex]\Delta h=h-(h_f+h')[/tex]
[tex]\Delta h=250-(240.1+9.8165)[/tex]
[tex]\Delta h=0.0835\ m[/tex]
