Meeting Points of Two Balls Dropping and Throwing Upward

This article delves into a classic physics problem involving the simultaneous dropping and throwing upward of two balls. Understanding this scenario not only serves as an excellent example of applying physics principles but also provides insight into the practical applications of mathematical equations.

Applying Physics Principles: Ball Collision

Consider a scenario where a ball is dropped from the top of a tower with a height of 100 meters, while at the same time, another ball is projected vertically upwards from the ground with an initial velocity of 25 m/s. Both balls are subject to the uniform acceleration due to gravity (g 9.81 m/s2).

Relative Velocity

Since both stones are subject to the same acceleration, the relative velocity between the two balls does not change. Initially, the relative velocity is 25 m/s, and it remains constant throughout their motion. The initial velocity of the first ball is 0 m/s, and the initial velocity of the second ball is 25 m/s.

Time to Meet

Using the equations for the motion of each ball, we can determine the time at which the two balls will meet. For the ball dropped from the tower:

Ball 1 (dropped from the tower):

[ y_1(t) 100 - frac{1}{2}g t^2 ]

For the ball projected upwards:

Ball 2 (thrown upwards):

[ y_2(t) 25t - frac{1}{2}g t^2 ]

Setting the positions of both balls equal at the meeting time:

[ 100 - frac{1}{2}g t^2 25t - frac{1}{2}g t^2 ]

Cancelling out the common terms:

[ 100 25t ]

Solving for ( t ):

[ t frac{100}{25} 4 , text{seconds} ]

Heights Calculation

Using ( t 4 , text{seconds} ), we can find the distance from the top of the tower where the two balls meet. Let's use the first ball's equation to find its position:

[ y_1(4) 100 - frac{1}{2}g (4)^2 ]

Calculating ( 4^2 16 ):

[ y_1(4) 100 - frac{1}{2} times 9.81 times 16 ]

[ y_1(4) 100 - 78.48 ]

[ y_1(4) 21.52 , text{meters} ]

Conclusion

The distance from the top of the tower at which the two balls meet is approximately 21.52 meters. This problem exemplifies the practical application of physics principles, particularly the use of kinematic equations under the influence of gravity.

Mathematical Model of Ball Motion

The scenario can be described by a system of two equations:

Ball 1 (dropped from the tower):

[ y_1(t) H - frac{1}{2}gt^2 ]

Ball 2 (thrown upwards):

[ y_2(t) v_0 t - frac{1}{2}gt^2 ]

By setting these equal at the meeting time, we can solve for the time ( t ) and then find the distance from the top of the tower using one of the equations.

Example Calculation

For a specific example, given ( H 100 , text{meters} ) and ( v_0 25 , text{m/s} ), we can calculate the exact time and position where the balls meet:

[ t_0 frac{H}{v_0} frac{100}{25} 4 , text{seconds} ]

[ h_0 100 - frac{1}{2}g (4)^2 ]

[ h_0 100 - frac{1}{2} times 9.81 times 16 ]

[ h_0 21.52 , text{meters} ]

This solution confirms that the balls will meet at a distance of 21.52 meters from the top of the tower after 4 seconds of motion.

Practical Applications

The problem discussed here finds relevance in various real-world applications. For instance, understanding the motion of falling objects and projectiles is crucial in disciplines like engineering, sports, and even telecommunications. The principles involved in this scenario can help in designing safe structures, optimizing game strategies, and improving existing technologies.