# 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.