Velocity of Balls Experiment
In the given experiment, one ball is dropped from a height of 12.3 m, while the other ball is thrown from a height of 25.0 m with an initial velocity of 6.35 m/s. We can use the equation v^2 = u^2 + 2as to calculate the final velocity of each ball as it strikes the ground.
For the ball dropped from 12.3 m, the initial velocity (u) is 0 m/s, the acceleration due to gravity (a) is -9.8 m/s^2, and the distance traveled (s) is 12.3 m. Plugging these values into the equation, we get v^2 = 0 + 2(-9.8)(12.3), which simplifies to v^2 = -239.16. Since we only consider positive velocities, the velocity of this ball as it strikes the ground is approximately 15.47 m/s.
For the ball thrown from 25.0 m with an initial velocity of 6.35 m/s, the acceleration due to gravity remains -9.8 m/s^2, and the distance traveled is 25.0 m. Plugging in these values, we get v^2 = (6.35)^2 + 2(-9.8)(25.0), which simplifies to v^2 = 286.75. Therefore, the velocity of this ball as it strikes the ground is approximately 16.94 m/s.
From the calculations, we can see that the two balls have different velocities when they hit the ground due to their initial conditions. This experiment demonstrates the effect of height and initial velocity on the final velocity of an object when dropped or thrown.