A softball having a mass of 0.30 kg is pitched horizontally at 120 km/h . By the time it reaches the plate, it may have slowed by 15 % .
The correct answer and explanation is:
Correct Answer:
The final speed of the softball is reduced by 15% from its initial speed.
Explanation:
To solve this problem, we need to calculate the final speed of the softball after it has slowed by 15% from its initial speed.
- Convert the initial speed into meters per second:
The initial speed of the softball is given as 120 km/h. To convert this to meters per second (m/s), we use the conversion factor: 1 km/h=1000 m3600 s=518 m/s1 \, \text{km/h} = \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{5}{18} \, \text{m/s} So: 120 km/h=120×518 m/s=33.33 m/s120 \, \text{km/h} = 120 \times \frac{5}{18} \, \text{m/s} = 33.33 \, \text{m/s} Therefore, the initial velocity of the softball is v0=33.33 m/sv_0 = 33.33 \, \text{m/s}. - Calculate the 15% reduction:
The problem states that the ball slows by 15% by the time it reaches the plate. This means it retains 85% of its original speed. The final speed vfv_f is: vf=0.85×v0=0.85×33.33 m/s=28.33 m/sv_f = 0.85 \times v_0 = 0.85 \times 33.33 \, \text{m/s} = 28.33 \, \text{m/s} So, the final speed of the softball when it reaches the plate is 28.33 m/s28.33 \, \text{m/s}.
Conclusion:
The softball’s final speed is 28.33 m/s28.33 \, \text{m/s} after slowing by 15% from its original speed of 33.33 m/s. The mass of the ball (0.30 kg) and the horizontal nature of the pitch are not directly required to calculate the speed change, but the mass would be relevant if the force or energy change were being considered.