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From the top of the tower of height 400m ,a ball is dropped by a man ,simultaneously from the Base of the tower, another ball is thrown up with velocity 50m/s ,at what distance will they meet from the Base of the tower ?option is (a) 100m (b) 80m (c) 320m (d) 240m?
Verified Answer
From the top of the tower of height 400m ,a ball is dropped by a man ,...
A ball dropped from the roof of height 400m Let after t time it meets with another ball .
displacement covered by ball during this time ( S1 ) = ut + 1/2at^2
= 0x t + 1/2 x 10 x t^2
= 5t^2
another ball is thrown from the surface of tower with speed 50m/sec .
then, displacement covered by another ball (S2) = ut +1/2at^2
= 50t - 1/2x 10t^2
=50t -5t^2
for collision,
S1 + S2 = 400
5t^2 + 50t - 5t^2 = 400
50t = 400
t = 8sec
hence, body collide in 8sec.
then, displacement covered by first body during this time= 5 x 64 = 320m
e.g collision occurs 80m height from the surface of tower .
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Most Upvoted Answer
From the top of the tower of height 400m ,a ball is dropped by a man ,...
Given:
- Height of the tower = 400m
- Initial velocity of the ball thrown up = 50m/s
To find:
- Distance at which the two balls will meet from the base of the tower
Assumptions:
- Neglect air resistance
Analysis:
Let's analyze the motion of the two balls separately to find the distance at which they meet.
Ball A (Dropped ball):
- Initial velocity (u) = 0 (since it is dropped)
- Acceleration (a) = 9.8 m/s² (acceleration due to gravity)
- Distance covered (s) = unknown
- Time taken (t) = unknown
Using the equation of motion, s = ut + 1/2at², we can calculate the time taken by the dropped ball to cover the height of the tower.
s = 400m (height of the tower)
u = 0
a = 9.8 m/s²
400 = 0*t + (1/2)*9.8*t²
400 = 4.9t²
t² = 400/4.9
t² ≈ 81.63
t ≈ √(81.63)
t ≈ 9.04s (approx.)
Hence, the dropped ball takes approximately 9.04 seconds to cover the height of the tower.
Ball B (Thrown ball):
- Initial velocity (u) = 50m/s
- Acceleration (a) = -9.8 m/s² (opposite to the direction of motion)
- Distance covered (s) = unknown
- Time taken (t) = 9.04s (as calculated above)
Using the equation of motion, s = ut + 1/2at², we can calculate the distance covered by the thrown ball in 9.04 seconds.
s = ?
u = 50m/s
a = -9.8 m/s²
t = 9.04s
s = 50*9.04 + (1/2)*(-9.8)*(9.04)²
s ≈ 452 + (-39.8)*(81.63)
s ≈ 452 - 3241.37
s ≈ -2789.37m (approx.)
The distance covered by the thrown ball in 9.04 seconds is approximately -2789.37 meters. Since distance cannot be negative, we take the absolute value.
|s| = 2789.37m ≈ 2789m
Answer:
The two balls will meet at a distance of approximately 2789 meters from the base of the tower.
- Height of the tower = 400m
- Initial velocity of the ball thrown up = 50m/s
To find:
- Distance at which the two balls will meet from the base of the tower
Assumptions:
- Neglect air resistance
Analysis:
Let's analyze the motion of the two balls separately to find the distance at which they meet.
Ball A (Dropped ball):
- Initial velocity (u) = 0 (since it is dropped)
- Acceleration (a) = 9.8 m/s² (acceleration due to gravity)
- Distance covered (s) = unknown
- Time taken (t) = unknown
Using the equation of motion, s = ut + 1/2at², we can calculate the time taken by the dropped ball to cover the height of the tower.
s = 400m (height of the tower)
u = 0
a = 9.8 m/s²
400 = 0*t + (1/2)*9.8*t²
400 = 4.9t²
t² = 400/4.9
t² ≈ 81.63
t ≈ √(81.63)
t ≈ 9.04s (approx.)
Hence, the dropped ball takes approximately 9.04 seconds to cover the height of the tower.
Ball B (Thrown ball):
- Initial velocity (u) = 50m/s
- Acceleration (a) = -9.8 m/s² (opposite to the direction of motion)
- Distance covered (s) = unknown
- Time taken (t) = 9.04s (as calculated above)
Using the equation of motion, s = ut + 1/2at², we can calculate the distance covered by the thrown ball in 9.04 seconds.
s = ?
u = 50m/s
a = -9.8 m/s²
t = 9.04s
s = 50*9.04 + (1/2)*(-9.8)*(9.04)²
s ≈ 452 + (-39.8)*(81.63)
s ≈ 452 - 3241.37
s ≈ -2789.37m (approx.)
The distance covered by the thrown ball in 9.04 seconds is approximately -2789.37 meters. Since distance cannot be negative, we take the absolute value.
|s| = 2789.37m ≈ 2789m
Answer:
The two balls will meet at a distance of approximately 2789 meters from the base of the tower.
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From the top of the tower of height 400m ,a ball is dropped by a man ,simultaneously from the Base of the tower, another ball is thrown up with velocity 50m/s ,at what distance will they meet from the Base of the tower ?option is (a) 100m (b) 80m (c) 320m (d) 240m?
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From the top of the tower of height 400m ,a ball is dropped by a man ,simultaneously from the Base of the tower, another ball is thrown up with velocity 50m/s ,at what distance will they meet from the Base of the tower ?option is (a) 100m (b) 80m (c) 320m (d) 240m? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about From the top of the tower of height 400m ,a ball is dropped by a man ,simultaneously from the Base of the tower, another ball is thrown up with velocity 50m/s ,at what distance will they meet from the Base of the tower ?option is (a) 100m (b) 80m (c) 320m (d) 240m? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for From the top of the tower of height 400m ,a ball is dropped by a man ,simultaneously from the Base of the tower, another ball is thrown up with velocity 50m/s ,at what distance will they meet from the Base of the tower ?option is (a) 100m (b) 80m (c) 320m (d) 240m?.
From the top of the tower of height 400m ,a ball is dropped by a man ,simultaneously from the Base of the tower, another ball is thrown up with velocity 50m/s ,at what distance will they meet from the Base of the tower ?option is (a) 100m (b) 80m (c) 320m (d) 240m? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about From the top of the tower of height 400m ,a ball is dropped by a man ,simultaneously from the Base of the tower, another ball is thrown up with velocity 50m/s ,at what distance will they meet from the Base of the tower ?option is (a) 100m (b) 80m (c) 320m (d) 240m? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for From the top of the tower of height 400m ,a ball is dropped by a man ,simultaneously from the Base of the tower, another ball is thrown up with velocity 50m/s ,at what distance will they meet from the Base of the tower ?option is (a) 100m (b) 80m (c) 320m (d) 240m?.
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