Meeting Height of Two Balls: A Physics Challenge

Two common scenarios in basic physics involve dropping and throwing objects. Understanding the motion and meeting height of objects in these scenarios is a fundamental concept. This article explores, in detail, the calculation involved when a ball is dropped from a height of 400 meters, while simultaneously another ball is thrown upward with a velocity of 50 meters per second.

Theoretical Background

The motion of an object in the absence of air resistance follows the laws of physics, particularly the principles of acceleration due to gravity. This article focuses on a problem where the first ball is dropped from a height of 50 meters, while the second ball is thrown upward with an initial velocity of 50 meters per second. The meeting height of these two balls is to be determined.

Equations and Calculations

Let's denote the height of the first ball, which is dropped, as ( s_1 ), and the height of the second ball, which is thrown upward, as ( s_2 ).

First Ball: Dropped from a Height

The initial velocity ( u_1 ) of the first ball is 0 m/s. The acceleration due to gravity ( g ) is -10 m/s2. The distance ( s_1 ) traveled by the first ball can be calculated using the formula: t

s1  u1t   1/2 * g * t^2

Substituting the values, we get:s1 0 1/2 * (-10) * t^2 s1 -5t^2

Second Ball: Thrown Upward

The initial velocity ( u_2 ) of the second ball is 50 m/s. The distance ( s_2 ) traveled by the second ball can be calculated using the same formula:s2 u2t 1/2 * g * t^2 Substituting the values, we get:s2 50t 1/2 * (-10) * t^2 s2 50t - 5t^2

Meeting Height Calculation

The total distance traveled by both stones when they meet will be equal to the initial height of the first stone, which is 50 meters. So, we can write:

s1   s2  50

Substituting the values of ( s_1 ) and ( s_2 ) in the equation, we get:

-5t^2   50t - 5t^2  50

Simplifying the equation, we get:

-10t^2   50t - 50  0

Dividing both sides by -10, we get:

t^2 - 5t   5  0

Solving the quadratic equation, we get:

t  5 ± sqrt(5^2 - 4*5) / 2
  t  5 ± sqrt(25 - 20) / 2
  t  5 ± sqrt(5) / 2
  t  5 ± 1.118 / 2

Since time cannot be negative, we take the positive value:

t  5   1.118 / 2 ≈ 3.62 seconds

NB: This is the time at which the two balls meet.

Meeting Height

Now, we can calculate the height at which the two balls meet by substituting the value of ( t ) in either ( s_1 ) or ( s_2 ).

Lets use ( s_1 ):

s1  -5t^2
s1  -5 * 3.62^2
s1 ≈ -65.61 meters

Since the height cannot be negative, we ignore the negative sign. Therefore, the two balls meet at a height of approximately 65.61 meters from the ground.

Conclusion and Additional Scenarios

The same principles can be applied to other scenarios. For example, if a ball is dropped from a height of 10 meters, the meeting height can be calculated using the formula:

d  1/2 * g * t^2   v*t   c
  d  -4.9t^2   10t   10
  d  10 - 4.9t^2

When the ball is thrown from the ground, the motion is:

d  1/2 * g * t^2   v*t   c
  d  -4.9t^2   10t - 10
  d  10t - 4.9t^2

They meet when ( d d ) so:

10 - 4.9t^2  10t - 4.9t^2
10t  10
  t  1
  d  10 - 4.9 * 1^2
  d  5.1 meters above the ground

If balls are dropped or thrown from different initial heights, the meeting height depends on their initial velocities and the distance they need to travel. They may never meet if their paths do not intersect.

Keywords: physics problem, dropping ball, thrown ball, meeting height, velocity, acceleration