A pitcher throws a 0.192-kg baseball, and it approaches the bat at a speed of 43.2 m/s. The bat does 69.8 J of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is 32.6 m above the point of impact.
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To determine the speed of the ball after it leaves the bat, we can use the principle of conservation of mechanical energy.
The mechanical energy of an object is the sum of its kinetic energy (KE) and potential energy (PE). In this case, we assume that the only forces acting on the ball are gravity and the work done by the bat.
Given:
Mass of the baseball (m) = 0.192 kg
Initial speed of the ball (Vi) = 43.2 m/s
Work done by the bat (W) = 69.8 J
Height above the point of impact (h) = 32.6 m
The initial mechanical energy (Ei) of the ball is the sum of its kinetic energy and potential energy:
Ei = KEi + PEi
KEi = 0.5 * m * Vi^2 (Equation 1)
PEi = m * g * h (Equation 2)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, we can find the final speed of the ball (Vf) using the conservation of mechanical energy:
Ei + W = KEf + PEf
Rearranging the equation and substituting the values:
KEf = Ei + W - PEf
KEf = KEi + W - PEf
Substituting the values from Equations 1 and 2:
0.5 * m * Vi^2 = 0.5 * m * Vf^2 + m * g * h
Simplifying the equation, we have:
Vi^2 = Vf^2 + 2 * g * h (Equation 3)
Now, we can solve for Vf. Rearranging Equation 3, we have:
Vf^2 = Vi^2 - 2 * g * h
Substituting the known values:
Vf^2 = (43.2 m/s)^2 - 2 * (9.8 m/s^2) * (32.6 m)
Calculating this expression, we find:
Vf^2 ≈ 935.424 m^2/s^2
To find Vf, we take the square root of both sides:
Vf ≈ √935.424 m^2/s^2
Vf ≈ 30.60 m/s
Therefore, the speed of the ball after leaving the bat and 32.6 m above the point of impact is approximately 30.60 m/s.