IDNLearn.com is your go-to platform for finding accurate and reliable answers. Our experts provide accurate and detailed responses to help you navigate any topic or issue with confidence.

Two balls are dropped from rest and allowed to fall. If one ball is allowed to fall for 1 s and the other for 3 s compare the distances they have fallen.

Sagot :

The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.

The given parameters;

  • time of fall of the first ball, t = 1 s
  • time of fall of the second ball, t = 3 s

The distance traveled by each ball is calculated using the second equation of motion as shown below.

The distance traveled by the first ball is calculated as follows;

[tex]h = u_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 1^2)\\\\h = 4.9 \ m[/tex]

The distance traveled by the second ball is calculated as follows;

[tex]h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 3^2)\\\\h = 44.1\ m[/tex]

Thus, the second ball traveled a greater distance because it spent more time in motion.

Learn more here:https://brainly.com/question/5868480