JEE Main 2021, 25 July Shift-II
LEVELJEE Main
Visualized Solution
The Sigma Insight: Motion in a Straight Line
The Setup
A Rising Balloon
Imagine you are standing on the ground, watching a hot air balloon steadily rising into the sky. The problem tells us that the balloon is moving upwards with a uniform velocity of . At the exact moment the balloon reaches a height of above the ground, a small object is dropped from it.
Our goal is to find out exactly how high the balloon is from the ground at the very instant this dropped object finally strikes the earth. To do this, we need to track two separate journeys: the downward fall of the object and the continuous upward rise of the balloon.
The Drop
Inertia in Action
Here is where many students make a classic mistake. When the object is "dropped," you might instinctively think its initial velocity is zero. But remember the concept of inertia of motion! Because the object was inside a balloon moving upwards at , the object itself was also moving upwards at .
Therefore, the moment it is released, it doesn't just fall straight down. It starts with an initial upward velocity of . Once it leaves the balloon, the only force acting on it is gravity, which pulls it downwards with an acceleration of .
The Math
Tracking the Object
We need to find out how long the object takes to hit the ground. Let's use the second equation of motion:
Let's set our sign convention carefully. We will take the upward direction as positive. The object starts at a height of and ends up on the ground. This means its total displacement is downwards, so .
Substituting our known values into the equation:
Simplifying this, we get:
Dividing the entire equation by to make the math cleaner, we arrive at a neat quadratic equation:
Factoring this quadratic equation gives us:
This yields two possible times: or . Since time cannot be negative in this physical context, we conclude that the object takes exactly to hit the ground.
The Balloon's Journey
While the object was busy executing its parabolic fall, what was the balloon doing? The problem states that the balloon moves with a uniform velocity. This means it has zero acceleration and simply continues to rise at for those same .
We can calculate the additional distance the balloon covers during this time using the simple formula for uniform motion:
The Final Tally
Finally, to find the balloon's total height from the ground when the object strikes the earth, we simply add this newly covered distance to its initial height.
The balloon is exactly high when the object hits the ground. The correct option is (c).
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