Answer:
The speed of the resistive force is 42.426 m/s
Explanation:
Given;
mass of skydiver, m = 75 kg
terminal velocity, [tex]V_T = 60 \ m/s[/tex]
The resistive force on the skydiver is known as drag force.
Drag force is directly proportional to square of terminal velocity.
[tex]F_D = kV_T^2[/tex]
Where;
k is a constant
[tex]k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}[/tex]
When the new drag force is half of the original drag force;
[tex]F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}= \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}= 0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s[/tex]
Therefore, the speed of the resistive force is 42.426 m/s
At terminal speed, the speed of the resistive force will be:
"42.426 m/s".
According to the question,
Skydriver's mass, m = 75 kg
Terminal velocity, [tex]V_T[/tex] = 60 m/s
Constant = k
We know the relation,
→ [tex]F_D[/tex] = k[tex]V_T^2[/tex]
here, k = [tex]\frac{F_D_1}{V_T_1^2} = \frac{F_D_2}{V_T_2^2}[/tex]
Now,
[tex]F_D_2[/tex] = [tex]\frac{F_D_1}{2}[/tex]
[tex]\frac{F_D_1}{V_T_1^2}= \frac{F_D_2}{V_T_2^2}[/tex]
[tex]\frac{1}{V_T_1^2} = \frac{1}{2V_T_2^2}[/tex]
By applying cross-multiplication,
[tex]V_T_2^2 = \sqrt{\frac{V_T_1^2}{2} }[/tex]
By substituting the above values,
[tex]V_T_2[/tex] = 0.7071 ([tex]V_T_1[/tex])
= 0.7071 × 60
= 42.426 m/s
Thus the above response is correct.
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