Meeting Points of Two Balls: A Physics Problem Explained

Understanding the motion and meeting point of two balls under gravity is a classic physics problem that involves the application of kinematic equations. Let's break down the scenario where a ball is dropped from the top of a tower, and another ball is thrown downward a few seconds later with a specific velocity. We will solve for the height at which the two balls meet.

Problem Setup

A ball is dropped from the top of a tower at a height of 50 meters. After 2 seconds, another ball is thrown vertically downward with a speed of 40 meters per second. We need to find at what height the two balls meet.

Equations of Motion

The motion of the balls can be described using the following equations:

Ball 1: Dropped from 50 meters

Initial height: h1 50 meters Initial velocity: u1 0 m/s (since it is dropped) Acceleration due to gravity: a1 9.81 m/s2 Time: t1 (time after the first ball is dropped)

The height of Ball 1 after t1 seconds is given by the equation:

Equation for Ball 1:

h1t1 h1 - 1/2 g t12

Substituting the values:

h1t1 50 - 1/2 (9.81) t12

Ball 2: Thrown downward

Initial height: h2 50 meters Initial velocity: u2 40 m/s Acceleration due to gravity: a2 9.81 m/s2 Time: t2 t1 - 2 seconds

The height of Ball 2 after t2 seconds is given by the equation:

Equation for Ball 2:

h2t2 h2 - u2 t2 - 1/2 g t22

Substituting t2 t1 - 2 into the equation:

h2t2 50 - 40 (t1 - 2) - 1/2 (9.81) (t1 - 2)2

Setting the heights equal:

50 - 1/2 (9.81) t12 50 - 40 (t1 - 2) - 1/2 (9.81) (t1 - 2)2

Simplifying the equation:

-1/2 (9.81) t12 -40 t1 80 - 1/2 (9.81) (t1 - 2)2

Expanding and rearranging:

-1/2 (9.81) t12 1/2 (9.81) t12 40 t1 - 80 -1/2 (9.81) (t1 - 2)2

40 t1 - 80 -1/2 (9.81) (t12 - 4 t1 4)

Multiplying through by -2:

-80 t1 160 -9.81 t12 19.62 t1 - 19.62

Rearranging for t1：

9.81 t12 - 80 t1 - 19.62 19.62 160

9.81 t12 - 80 t1 - 144 0

Solving the quadratic equation:

t1 (80 ± sqrt((80)2 - 4 (9.81) (-144))) / 2 (9.81)

t1 (80 ± sqrt(6400 5595.64)) / 19.62

t1 (80 ± sqrt(11995.64)) / 19.62

t1 (80 ± 109.52) / 19.62

t1 189.52 / 19.62 or -29.52 / 19.62

t1 4.04 seconds (ignoring negative solution)

Final Height Calculation

Substituting t1 4.04 seconds back into the height equation for Ball 1:

h1t1 50 - 1/2 (9.81) (4.04)2

Calculating h1t1

h1t1 50 - 0.5 (9.81) (16.32)

h1t1 50 - 80.03 ≈ 49.84 meters

Therefore, the two balls meet at approximately 49.84 meters above the ground.