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A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity 25 ms–1. Then the distance from the top of the tower, at which the two balls meet is
- a)68.4 m
- b)48.4 m
- c)18.4 m
- d)78.4 m
Correct answer is option 'D'. Can you explain this answer?
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A ball is dropped from the top of a tower of height 100 m and at the ...
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A ball is dropped from the top of a tower of height 100 m and at the ...
Given:
Height of the tower, h = 100 m
Initial velocity of the ball projected upwards, u = 25 m/s
Acceleration due to gravity, g = 9.8 m/s²
To find:
Distance from the top of the tower, at which the two balls meet
Explanation:
Let's first calculate the time taken by each ball to reach the meeting point.
For the ball dropped from the top of the tower:
Using the equation of motion,
h = ut + 1/2 gt²
where h = 100 m, u = 0 m/s (as it is dropped), g = 9.8 m/s²
On substituting the values, we get
100 = 0 × t + 1/2 × 9.8 × t²
Simplifying, we get
t² = 100/4.9
t = √(100/4.9) = 4.52 s
For the ball projected upwards from ground:
Using the equation of motion,
v = u + gt
where v = 0 m/s (at the highest point), u = 25 m/s, g = 9.8 m/s²
On substituting the values, we get
0 = 25 – 9.8 × t
t = 25/9.8 = 2.55 s
Now, let's calculate the distance from the top of the tower, at which the two balls meet.
Distance travelled by the ball dropped from the top of the tower in 4.52 s:
Using the equation of motion,
s = ut + 1/2 gt²
where u = 0 m/s, g = 9.8 m/s², t = 4.52 s
On substituting the values, we get
s = 0 × 4.52 + 1/2 × 9.8 × (4.52)² = 100 m
Distance travelled by the ball projected upwards from ground in 2.55 s:
Using the equation of motion,
s = ut + 1/2 gt²
where u = 25 m/s, g = 9.8 m/s², t = 2.55 s
On substituting the values, we get
s = 25 × 2.55 + 1/2 × 9.8 × (2.55)² = 78.4 m
Therefore, the distance from the top of the tower, at which the two balls meet is 78.4 m.
Hence, the correct answer is option D.
Height of the tower, h = 100 m
Initial velocity of the ball projected upwards, u = 25 m/s
Acceleration due to gravity, g = 9.8 m/s²
To find:
Distance from the top of the tower, at which the two balls meet
Explanation:
Let's first calculate the time taken by each ball to reach the meeting point.
For the ball dropped from the top of the tower:
Using the equation of motion,
h = ut + 1/2 gt²
where h = 100 m, u = 0 m/s (as it is dropped), g = 9.8 m/s²
On substituting the values, we get
100 = 0 × t + 1/2 × 9.8 × t²
Simplifying, we get
t² = 100/4.9
t = √(100/4.9) = 4.52 s
For the ball projected upwards from ground:
Using the equation of motion,
v = u + gt
where v = 0 m/s (at the highest point), u = 25 m/s, g = 9.8 m/s²
On substituting the values, we get
0 = 25 – 9.8 × t
t = 25/9.8 = 2.55 s
Now, let's calculate the distance from the top of the tower, at which the two balls meet.
Distance travelled by the ball dropped from the top of the tower in 4.52 s:
Using the equation of motion,
s = ut + 1/2 gt²
where u = 0 m/s, g = 9.8 m/s², t = 4.52 s
On substituting the values, we get
s = 0 × 4.52 + 1/2 × 9.8 × (4.52)² = 100 m
Distance travelled by the ball projected upwards from ground in 2.55 s:
Using the equation of motion,
s = ut + 1/2 gt²
where u = 25 m/s, g = 9.8 m/s², t = 2.55 s
On substituting the values, we get
s = 25 × 2.55 + 1/2 × 9.8 × (2.55)² = 78.4 m
Therefore, the distance from the top of the tower, at which the two balls meet is 78.4 m.
Hence, the correct answer is option D.
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A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity 25 ms–1. Then the distance from the top of the tower, at which the two balls meet isa)68.4 mb)48.4 mc)18.4 md)78.4 mCorrect answer is option 'D'. Can you explain this answer?
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A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity 25 ms–1. Then the distance from the top of the tower, at which the two balls meet isa)68.4 mb)48.4 mc)18.4 md)78.4 mCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity 25 ms–1. Then the distance from the top of the tower, at which the two balls meet isa)68.4 mb)48.4 mc)18.4 md)78.4 mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity 25 ms–1. Then the distance from the top of the tower, at which the two balls meet isa)68.4 mb)48.4 mc)18.4 md)78.4 mCorrect answer is option 'D'. Can you explain this answer?.
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