Answer:
The speed of the ball is 42.5 m/s
Explanation:
The initial kinetic energy of the ball is:
[tex]K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2[/tex]= 85.75 J
The speed of the ball after leaving the bat is:
[tex]K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}[/tex]
V=47.92 m/s
Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:
[tex]V_f^2-V^2=-2gh[/tex]
[tex]V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m[/tex]
[tex]V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2[/tex]
[tex]V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}[/tex]
[tex]V_f=42.5m/s[/tex]
