Answer in Mechanics | Relativity for Joey #108647

As per the given question,

Mass of the stakeholder (m)=62.7 kg

Initial height “(h_1)=4.55 m”

Final height “(h_2)=5.7 m”

Final speed of the stakeholder (v)=6 m/sec

Work done by the friction =120J

Let the initial speed of the stakeholder is (u),

Now applying the conservation of energy,

“mgh_1+ dfrac{mu^2}{2}+W_{fs}=mgh_2+dfrac{mv^2}{2}”

Now substituting the values,

“62.7times 9.8times 4.55+ dfrac{62.7times u^2}{2}+120=62.7times 9.8times 5.7+dfrac{62.7times6^2}{2}”

“Rightarrow dfrac{62.7times u^2}{2}=3502.422+1128.6-120-2795.793”

“Rightarrow u^2=dfrac{1715.229times2}{62.7}”

“Rightarrow u=sqrt{54.71}”

“u=7.4 m/sec”