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Answer :
The initial kinetic energy of the ball is 12,330 Joules when it is launched at 54 m/s at an angle of 67 degrees.
The initial kinetic energy of the ball can be found using the equation
[tex]KE = (1/2)*m*v^2[/tex]
where m is the mass of the ball and v is the velocity of the ball.
In this problem, given the mass of ball (m) = 7.0 kg
The velocity at which the ball is launched (v) = 54 m/s.
The angle at which the ball is launched = 67°
The velocity along the horizontal = Vx = Vcos67° = 21.09m/s
The velocity along vertical direction Vy = Vsin67° = 49.7m/s
The resultant velocity[tex]V = \sqrt{ Vx^2 + Vy^2} =\sqrt{(21.09)^2 + (49.7)^2}[/tex]
V = 53.98 ≈ 54m/s
Therefore, the initial kinetic energy of the ball is:
[tex]KE = (1/2)*7.0 kg*(54 m/s)^2[/tex]
[tex]KE = (1/2)*7.0 kg*2916 m^2/s^2[/tex]
[tex]KE = 12,330 kg m^2/s^2[/tex]
KE = 12,330 Joules
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