Answer:
a)43.8 m/s
b)103.54 m/s
Explanation:
Work energy theorem
Δ[tex]E_{k}=[/tex] Δ[tex]E_{g} [/tex]
[tex]E_{k2} - E_{k1}=-(E_{g2} - E_{g1})\\\\ \frac{1}{2} mv^{2}_{2}- \frac{1}{2} mv^{2}_{1}+ mgh_{2} - mgh_{1} = 0[/tex]
a) In this part v1=0 and h2=0
[tex]\frac{1}{2} mv^{2}_{2}- \frac{1}{2} mv^{2}_{1}+ mgh_{2}- mgh_{1} = 0\\\\\\\\\v_{2} =\sqrt{2gh_{1}} = \sqrt{2*9.8*98} =43.8 m/s[/tex]=0
b) in this part v2=0, h1= 0, and h2= 545m
[tex]\frac{1}{2} v^{2}_{1}= gh_{2} \\v_{1} =\sqrt{2gh_{2}}=103.54m/s[/tex]
