Answer:
0.02867 m
1.722 m
Explanation:
x = Compression of net
h = Height of jump
g = Acceleration due to gravity = 9.81 m/s²
The potential energy and the kinetic energy of the system is conserved
[tex]P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 70\times 9.81\frac{20+1.1}{1.1^2}\\\Rightarrow k=23949.3719\ N/m[/tex]
The spring constant of the net is 23949.3719 N
From Hooke's Law
[tex]F=kx\\\Rightarrow x=\dfrac{F}{k}\\\Rightarrow x=\frac{70\times 9.81}{23949.3719}\\\Rightarrow x=0.02867\ m[/tex]
The net would stretch 0.02867 m
If h = 50 m
From energy conservation
[tex]70\times 9.81\times (50+x)=\frac{1}{2}23949.3719x^2\\\Rightarrow 11974.68595x^2=686.7(50+x)\\\Rightarrow 50+x=17.43801x^2\\\Rightarrow 17.43801x^2-x-50=0[/tex]
Solving the above equation we get
[tex]x=\frac{-\left(-1\right)+\sqrt{\left(-1\right)^2-4\cdot \:17.43801\left(-50\right)}}{2\cdot \:17.43801}, \frac{-\left(-1\right)-\sqrt{\left(-1\right)^2-4\cdot \:17.43801\left(-50\right)}}{2\cdot \:17.43801}\\\Rightarrow x=1.722, -1.6[/tex]
The compression of the net is 1.722 m
