Answer:
[tex]24.9\ \text{m/s}[/tex]
[tex]206.7\ \text{m}[/tex]
Explanation:
m = Mass of skydiver = 80 kg
[tex]x_1[/tex] = Height for which the parachute is closed = 1000-200 = 800 m
[tex]x_2[/tex] = Height for which the parachute is open = 200 m
[tex]f_1[/tex] = Resistive force when parachute is closed = 50 N
[tex]f_2[/tex] = Resistive force when parachute is open = 3600 N
v = Velocity of skydiver on the ground
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
h = Height from which the skydiver jumps = 1000 m
The energy balance of the system will be
[tex]mgh-f_1x_1-f_2x_2=\dfrac{1}{2}mv^2\\\Rightarrow 80\times 9.81\times 1000-50\times 800-3600\times 200=\dfrac{1}{2}\times 80\times v^2\\\Rightarrow v=\sqrt{\dfrac{2(80\times 9.81\times 1000-50\times 800-3600\times 200)}{80}}\\\Rightarrow v=24.9\ \text{m/s}[/tex]
The velocity fo the skydiver when he lands will be [tex]24.9\ \text{m/s}[/tex]
x = Height where the person opens the parachute
v = 5 m/s
[tex]mgh-f_1x_1-f_2x_2=\dfrac{1}{2}mv^2\\\Rightarrow 80\times 9.81\times 1000-50\times (1000-x)-3600\times x=\dfrac{1}{2}\times 80\times 5^2\\\Rightarrow 80\times 9.81\times 1000-50000+50x-3600x=\dfrac{1}{2}\times 80\times 5^2\\\Rightarrow x=\dfrac{80\times 9.81\times 1000-50000-\dfrac{1}{2}\times 80\times 5^2}{3550}\\\Rightarrow x=206.7\ \text{m}[/tex]
The height at which the parachute is to be opened is [tex]206.7\ \text{m}[/tex]
