একটি বস্তকে 400 m উচ্চতা থেকে নিচে ছাড়া হলো এবং একই সময়ে একটি একটি বস্তকে 50 m/s বেগে নিচ থেকে খাড়া উপরে ছোড়া হলো। কত উচ্চতায় বস্ত দুইটি মিলিত হবে? $[g = 10 m/s^2]$

The correct answer to the question is 80 m because of how the two objects move due to the forces acting on them. Let's break it down:

Understanding the Motion:

    
• Object A is dropped from a height of 400 m. It starts with an initial velocity of 0 m/s and falls under the influence of gravity.
    
• Object B is thrown upward from the ground with an initial velocity of 50 m/s. As it goes up, it will slow down due to gravity acting against it.

Finding the Meeting Point:

    
• When both objects meet, the height of Object A (falling from 400 m) and Object B (going upwards) will be equal.
    
• Using the equations of motion for both objects, we can find a height where their paths intersect.

Calculation Overview:

    
• For Object A falling from 400 m, the distance fallen at time t is given by:         s = 400 - (1/2)gt² where g is approximately 10 m/s².
    
• For Object B thrown upwards:         s = 50t - (1/2)gt².

By equating the height of both objects at the time they meet, you'll find that when solved, it leads to the conclusion that they will meet at 80 m from the ground.

This explains why the answer is 80 m, as it is derived from the equations of motion for both objects and their starting conditions. Thus, among the options given, 80 m is indeed the point where they converge.

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