Question

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.

Answer 1

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;

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

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

$h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 3^2)\\\\h = 44.1\ m$

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

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