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A man in a balloon rising vertically with an acceleration of 4.9 m/sec2 releases a ball 2 sec after the balloon is let go from the ground. The greatest height above the ground reached by the ball is (g = 9.8 m/sec2)
- a)14.7 m
- b)19.6 m
- c)9.8 m
- d)24.5 m
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A man in a balloon rising vertically with an acceleration of 4.9 m/sec...
Understanding the Problem
The problem involves a balloon rising with a constant upward acceleration, and a ball being released from the balloon after 2 seconds. We need to calculate the maximum height reached by the ball after its release.
Key Information
- Acceleration of the balloon (a) = 4.9 m/s²
- Time before ball is released (t) = 2 s
- Acceleration due to gravity (g) = 9.8 m/s²
Calculating the Height of the Balloon
- The balloon starts from rest, so we use the formula for distance under constant acceleration:
Distance (h) = 0.5 * a * t²
- Substituting the values:
h = 0.5 * 4.9 m/s² * (2 s)²
h = 0.5 * 4.9 * 4
h = 9.8 m
Velocity of the Balloon at Release
- The velocity of the balloon at the time of release can be calculated as:
Velocity (v) = a * t
- Substituting the values:
v = 4.9 m/s² * 2 s
v = 9.8 m/s
Motion of the Ball After Release
- When the ball is released, it inherits the upward velocity of the balloon (9.8 m/s) but is now subject to the downward acceleration of gravity.
Calculating Maximum Height of the Ball
- The ball will rise until its velocity becomes zero. The time (t1) to reach maximum height can be found using:
v_final = v_initial - g * t1
0 = 9.8 m/s - 9.8 m/s² * t1
t1 = 1 s
- The additional height gained by the ball during this time is:
Height (h1) = v_initial * t1 - 0.5 * g * t1²
h1 = 9.8 m/s * 1 s - 0.5 * 9.8 m/s² * (1 s)²
h1 = 9.8 m - 4.9 m
h1 = 4.9 m
Total Height Above Ground
- The total height reached by the ball above ground is:
Total height = Height of the balloon + Height gained by the ball
Total height = 9.8 m + 4.9 m
Total height = 14.7 m
Thus, the greatest height above the ground reached by the ball is 14.7 m (option A).
The problem involves a balloon rising with a constant upward acceleration, and a ball being released from the balloon after 2 seconds. We need to calculate the maximum height reached by the ball after its release.
Key Information
- Acceleration of the balloon (a) = 4.9 m/s²
- Time before ball is released (t) = 2 s
- Acceleration due to gravity (g) = 9.8 m/s²
Calculating the Height of the Balloon
- The balloon starts from rest, so we use the formula for distance under constant acceleration:
Distance (h) = 0.5 * a * t²
- Substituting the values:
h = 0.5 * 4.9 m/s² * (2 s)²
h = 0.5 * 4.9 * 4
h = 9.8 m
Velocity of the Balloon at Release
- The velocity of the balloon at the time of release can be calculated as:
Velocity (v) = a * t
- Substituting the values:
v = 4.9 m/s² * 2 s
v = 9.8 m/s
Motion of the Ball After Release
- When the ball is released, it inherits the upward velocity of the balloon (9.8 m/s) but is now subject to the downward acceleration of gravity.
Calculating Maximum Height of the Ball
- The ball will rise until its velocity becomes zero. The time (t1) to reach maximum height can be found using:
v_final = v_initial - g * t1
0 = 9.8 m/s - 9.8 m/s² * t1
t1 = 1 s
- The additional height gained by the ball during this time is:
Height (h1) = v_initial * t1 - 0.5 * g * t1²
h1 = 9.8 m/s * 1 s - 0.5 * 9.8 m/s² * (1 s)²
h1 = 9.8 m - 4.9 m
h1 = 4.9 m
Total Height Above Ground
- The total height reached by the ball above ground is:
Total height = Height of the balloon + Height gained by the ball
Total height = 9.8 m + 4.9 m
Total height = 14.7 m
Thus, the greatest height above the ground reached by the ball is 14.7 m (option A).
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Community Answer
A man in a balloon rising vertically with an acceleration of 4.9 m/sec...
Just before Releasing:
(a) Velocity of stone = Velocity of balloon
(b) Acceleration of stone = Acceleration of balloon
Just After Releasing:
(a) Velocity of stone = velocity of balloon
(b) Acceleration of stone = 9.8 m/s2 downwards
Height at which stone is released = H
=
Velocity with which stone is released = V
= 0 + 4.9 × 2 = 9.8 m/sec
∴ Hmax above ground level =
= 14.7 m
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A man in a balloon rising vertically with an acceleration of 4.9 m/sec2 releases a ball 2 sec after the balloon is let go from the ground. The greatest height above the ground reached by the ball is (g = 9.8 m/sec2)a)14.7 mb)19.6 mc)9.8 md)24.5 mCorrect answer is option 'A'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A man in a balloon rising vertically with an acceleration of 4.9 m/sec2 releases a ball 2 sec after the balloon is let go from the ground. The greatest height above the ground reached by the ball is (g = 9.8 m/sec2)a)14.7 mb)19.6 mc)9.8 md)24.5 mCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A man in a balloon rising vertically with an acceleration of 4.9 m/sec2 releases a ball 2 sec after the balloon is let go from the ground. The greatest height above the ground reached by the ball is (g = 9.8 m/sec2)a)14.7 mb)19.6 mc)9.8 md)24.5 mCorrect answer is option 'A'. Can you explain this answer?.
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