A 0.31-kg stone is held 1.1 m above the top edge of a water well and then dropped into it. The well has a depth of 4.5 m. (a) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone−Earth system before the stone is released? J (b) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone−Earth system when it reaches the bottom of the well? J (c) What is the change in gravitational potential energy of the system from release to reaching the bottom of the well?

Liquid. good luck : )

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.