A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 6.78 m/s at the bottom of the rise. Find the translational speed at the top. The image depicts a scenario involving a rolling ball on a track with an elevation. The diagram shows two positions of the ball:1. **Initial Position:** The ball is rolling along a horizontal track. Orange arrows indicate the direction of motion and rotation of the ball.2. **Elevated Position:** The ball moves to a higher section of the track. The elevation is marked as 0.760 meters from the lower horizontal level. Again, orange arrows highlight the direction of motion and rotation.The illustration is likely used to explain concepts such as energy conservation, rolling motion, and potential energy in physics. The elevation indicates a change in gravitational potential energy as the ball moves upward on the track.

Take potential energy at bottom to be zero as reference and Use energy conservation in the given question as follows. Kinetic energy at bottom + potential energy at bottom = kinetic energy at top + Potential energy at top 12mu2+0==12mv2+mghu2=v2+2gh Here u is the velocity at the bottom, v is the velocity at the top and h is the height Given values are as follows: u = 6.78 m/s , h = 0.76 m Therefore, u2=v2+2gh6.782=v2+29.80.76v=1.75668 m/s