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A parachutist's fall to Earth is determined by two opposing forces. A gravitational force of 605 N acts on the parachutist. After 3 s, she opens her parachute and experiences an air resistance of 689 N. At what speed (in m/s) is the parachutist falling after 10 s?

Respuesta :

Answer:15.8 m/s

Explanation:

Given

Gravitational Force [tex]F_g=605 N[/tex]

Air resistance [tex]F_d=689 N[/tex]

mass of Parachutist [tex]=\frac{605}{9.8}=61.73 kg[/tex]

Velocity after [tex]3 s[/tex]

[tex]v=u+at[/tex]

[tex]v=0+9.8\times 3[/tex]

[tex]v=29.4 m/s[/tex]

Now Parachutist opens the parachute

Net for on Parachutist

[tex]605-689=-84 [/tex]

[tex]84 N[/tex] drag force

therefore deceleration [tex]\frac{84}{61.73}=1.36 m/s^2[/tex]

velocity after 10 s is

[tex]v_1=v+at[/tex]

[tex]v_1=29.4-1.36\times 10[/tex]

[tex]v_1=15.8 m/s[/tex]

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