Answer:195 J
Explanation:
Given
mass of ball [tex]m=0.0570\ kg[/tex]
ball leaves the hand with [tex]u=15\ m/s[/tex]
maximum height reached by ball [tex]h=8\ m[/tex]
Initial Mechanical energy when ball just leaves the hand
[tex]M.E._1=(P.E.+K.E.)_1[/tex]
[tex]M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1[/tex]
considering hand to be datum so h_1=0[/tex]
so Potential energy at ground is zero
[tex]M.E._1=\frac{1}{2}\times m\times (15)^2[/tex]
[tex]M.E._1=6.41\ J[/tex]
Mechanical Energy at highest point
[tex](M.E.)_2=(P.E.+K.E.)_2[/tex]
at highest Point velocity is zero
[tex](M.E.)_2=mgh_2+0[/tex]
[tex](M.E.)_2=0.0570\times 9.8\times 8[/tex]
[tex](M.E.)_2=4.46\ J[/tex]
Decrease in Mechanical energy
[tex](M.E.)_1-(M.E.)_2=6.41-4.46[/tex]
[tex](M.E.)_1-(M.E.)_2=1.95\ J[/tex]