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A ball is dropped from the top of a 100 m high tower on a planet. In the last 1/2s before hitting the
ground, it covers a distance of 19 m. Acceleration due to gravity (in ms-2) near the surface on that
planet is
ground, it covers a distance of 19 m. Acceleration due to gravity (in ms-2) near the surface on that
planet is
- a)8
- b)2888
- c)4
- d)3
Correct answer is option 'A,B'. Can you explain this answer?
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Most Upvoted Answer
A ball is dropped from the top of a 100 m high tower on a planet. In t...
**Given data:**
- Height of the tower (h) = 100 m
- Distance covered in the last 1/2 seconds (d) = 19 m
**To find:**
- Acceleration due to gravity (g)
**Explanation:**
To solve this problem, we can use the equations of motion.
Let's consider the initial velocity (u) of the ball as 0 since it is being dropped from rest.
Using the equation of motion:
d = ut + 1/2 * a * t^2
where,
- d is the distance covered
- u is the initial velocity
- a is the acceleration
- t is the time
Substituting the given values:
19 = 0 + 1/2 * a * (1/2)^2
19 = 1/8 * a
Rearranging the equation:
a = 19 * 8 = 152 m/s^2
However, this is not the acceleration due to gravity. This is the net acceleration experienced by the ball during the last 1/2 second before hitting the ground.
Now, let's consider the motion of the ball from the top of the tower to the point where it covers a distance of 19m in the last 1/2 second.
Using the equation of motion:
h = ut + 1/2 * a * t^2
where,
- h is the height of the tower
- u is the initial velocity
- a is the acceleration
- t is the time
Substituting the given values:
100 = 0 + 1/2 * a * t^2
Simplifying the equation:
100 = 1/2 * a * (1/2)^2
100 = 1/8 * a
Rearranging the equation:
a = 100 * 8 = 800 m/s^2
Since the acceleration experienced by the ball during its fall from the top of the tower is equal to the acceleration due to gravity, we can conclude that the acceleration due to gravity on that planet is 800 m/s^2 or 8 m/s^2.
Therefore, the correct answer is option A) 8.
- Height of the tower (h) = 100 m
- Distance covered in the last 1/2 seconds (d) = 19 m
**To find:**
- Acceleration due to gravity (g)
**Explanation:**
To solve this problem, we can use the equations of motion.
Let's consider the initial velocity (u) of the ball as 0 since it is being dropped from rest.
Using the equation of motion:
d = ut + 1/2 * a * t^2
where,
- d is the distance covered
- u is the initial velocity
- a is the acceleration
- t is the time
Substituting the given values:
19 = 0 + 1/2 * a * (1/2)^2
19 = 1/8 * a
Rearranging the equation:
a = 19 * 8 = 152 m/s^2
However, this is not the acceleration due to gravity. This is the net acceleration experienced by the ball during the last 1/2 second before hitting the ground.
Now, let's consider the motion of the ball from the top of the tower to the point where it covers a distance of 19m in the last 1/2 second.
Using the equation of motion:
h = ut + 1/2 * a * t^2
where,
- h is the height of the tower
- u is the initial velocity
- a is the acceleration
- t is the time
Substituting the given values:
100 = 0 + 1/2 * a * t^2
Simplifying the equation:
100 = 1/2 * a * (1/2)^2
100 = 1/8 * a
Rearranging the equation:
a = 100 * 8 = 800 m/s^2
Since the acceleration experienced by the ball during its fall from the top of the tower is equal to the acceleration due to gravity, we can conclude that the acceleration due to gravity on that planet is 800 m/s^2 or 8 m/s^2.
Therefore, the correct answer is option A) 8.
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A ball is dropped from the top of a 100 m high tower on a planet. In the last 1/2s before hitting theground, it covers a distance of 19 m. Acceleration due to gravity (in ms-2) near the surface on thatplanet isa)8b)2888c)4d)3Correct answer is option 'A,B'. Can you explain this answer?
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A ball is dropped from the top of a 100 m high tower on a planet. In the last 1/2s before hitting theground, it covers a distance of 19 m. Acceleration due to gravity (in ms-2) near the surface on thatplanet isa)8b)2888c)4d)3Correct answer is option 'A,B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A ball is dropped from the top of a 100 m high tower on a planet. In the last 1/2s before hitting theground, it covers a distance of 19 m. Acceleration due to gravity (in ms-2) near the surface on thatplanet isa)8b)2888c)4d)3Correct answer is option 'A,B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is dropped from the top of a 100 m high tower on a planet. In the last 1/2s before hitting theground, it covers a distance of 19 m. Acceleration due to gravity (in ms-2) near the surface on thatplanet isa)8b)2888c)4d)3Correct answer is option 'A,B'. Can you explain this answer?.
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