Physics Problem: Calculating Distance Between Two Vertically Thrown Balls

Introduction

Two physics students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s; at the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. How far apart are the balls 0.8 s after they are thrown?

Solution

The two thrown balls, one upwards and one downwards from a balcony, given similar initial conditions and after 0.8 seconds of their motion, are calculated to be 6.3 meters apart from each other.

Physics Concept

The subject of this question is related to the Physics concept of free fall and motion. The problem is about the motion of two balls thrown from a balcony 19.6 m above the ground, one upwards and one downwards with the same initial speed of 14.7 m/s. To solve this problem, we need to calculate the distance each ball has moved after 0.8 s of their motion.

Calculation

Looking specifically at the ball that is thrown downwards, we use the equation: h = gt²/2 + vt. Here, g is the gravity constant, t is time, v is the initial speed, and h is the height. Given that we know g = 9.8 m/s², v = 14.7 m/s, and t = 0.8s, we can solve for h₁ to find that h₁ = (9.8m/s² * (0.8s)²/2) + 14.7m/s * 0.8s = 3.136 m + 11.76 m = 14.9 m.

The ball thrown upwards would have moved a distance of h₂ = (9.8m/s² * (0.8s)²/2) + 14.7m/s * 0.8s = 3.136 m - 11.76 m = -8.6m. But since the negative value denotes upward motion and the ball is moving downward as it passes the balcony, we take the absolute value to be 8.6 m.

So, the two balls are 14.9 m - 8.6 m = 6.3 meters apart 0.8 seconds after they are thrown.