1. **State the problem:** A tennis ball of mass $0.058$ kg is dropped from a height of $0.87$ m. It loses $0.15$ J of mechanical energy due to collisions during its first bounce. We need to find the height the ball reaches after the bounce. 2. **Relevant formula:** The gravitational potential energy (GPE) at height $h$ is given by: $$PE = mgh$$ where $m$ is mass, $g$ is acceleration due to gravity ($9.8$ m/s$^2$), and $h$ is height. 3. **Calculate initial potential energy:** $$PE_{initial} = 0.058 \times 9.8 \times 0.87 = 0.4945\, J$$ 4. **Energy after bounce:** Energy lost to thermal energy is $0.15$ J, so mechanical energy after bounce is: $$PE_{final} = PE_{initial} - 0.15 = 0.4945 - 0.15 = 0.3445\, J$$ 5. **Calculate final height:** Using $PE = mgh$ again: $$h_{final} = \frac{PE_{final}}{mg} = \frac{0.3445}{0.058 \times 9.8}$$ 6. **Simplify the fraction:** $$h_{final} = \frac{0.3445}{0.5684}$$ 7. **Final calculation:** $$h_{final} \approx 0.606\, m$$ **Answer:** The ball reaches approximately $0.61$ meters after its first bounce.