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A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with speed v. The two balls meet at t =18 s. What is the value of v? (take g = 10 m/s^2) (i) 75 m/s (ii) 55m/s (iii) 40m/s (iv) 60 m/s Please help bayya. correct answer is (i). Can you explain why?
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A ball is dropped from a high rise platform at t=0 starting from rest....
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A ball is dropped from a high rise platform at t=0 starting from rest....
Solution:
Given that,
g = 10 m/s^2
Time taken by first ball to reach the ground (t1) = 6 seconds
Time taken by second ball to reach the ground (t2) = 18 seconds
Let the initial velocity of the second ball be v0.
Now, we can use the equations of motion to find the distance covered by each ball.
For the first ball:
u = 0 (initial velocity)
a = g (acceleration due to gravity)
t = t1 = 6 s
Using s = ut + (1/2)at^2,
s1 = 0 + (1/2) × 10 × (6)^2
s1 = 180 m
For the second ball:
u = v0
a = g
t = t2 - t1 = 12 s
Using s = ut + (1/2)at^2,
s2 = v0 × 12 + (1/2) × 10 × (12)^2
s2 = 6v0 + 720
Now, since the two balls meet at 18 seconds, they cover the same distance.
So, s1 = s2
180 = 6v0 + 720
6v0 = -540
v0 = -90 m/s (which is not possible)
So, there must be some mistake in our calculation. Let's check our calculations.
We can use another equation of motion to find the velocity of the first ball just before the second ball is thrown.
For the first ball:
u = 0
a = g
t = t2 = 18 s
Using v = u + at,
v1 = 0 + 10 × 18
v1 = 180 m/s
Now, since the two balls meet, the velocity of the second ball when they meet must be equal to v1.
Using v = u + at,
v1 = v0 + 10 × 12
180 = v0 + 120
v0 = 60 m/s
Therefore, the value of v is 60 m/s, which is option (iv).
Final Answer: Option (iv) 60 m/s.
Given that,
g = 10 m/s^2
Time taken by first ball to reach the ground (t1) = 6 seconds
Time taken by second ball to reach the ground (t2) = 18 seconds
Let the initial velocity of the second ball be v0.
Now, we can use the equations of motion to find the distance covered by each ball.
For the first ball:
u = 0 (initial velocity)
a = g (acceleration due to gravity)
t = t1 = 6 s
Using s = ut + (1/2)at^2,
s1 = 0 + (1/2) × 10 × (6)^2
s1 = 180 m
For the second ball:
u = v0
a = g
t = t2 - t1 = 12 s
Using s = ut + (1/2)at^2,
s2 = v0 × 12 + (1/2) × 10 × (12)^2
s2 = 6v0 + 720
Now, since the two balls meet at 18 seconds, they cover the same distance.
So, s1 = s2
180 = 6v0 + 720
6v0 = -540
v0 = -90 m/s (which is not possible)
So, there must be some mistake in our calculation. Let's check our calculations.
We can use another equation of motion to find the velocity of the first ball just before the second ball is thrown.
For the first ball:
u = 0
a = g
t = t2 = 18 s
Using v = u + at,
v1 = 0 + 10 × 18
v1 = 180 m/s
Now, since the two balls meet, the velocity of the second ball when they meet must be equal to v1.
Using v = u + at,
v1 = v0 + 10 × 12
180 = v0 + 120
v0 = 60 m/s
Therefore, the value of v is 60 m/s, which is option (iv).
Final Answer: Option (iv) 60 m/s.
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A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with speed v. The two balls meet at t =18 s. What is the value of v? (take g = 10 m/s^2) (i) 75 m/s (ii) 55m/s (iii) 40m/s (iv) 60 m/s Please help bayya. correct answer is (i). Can you explain why?
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A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with speed v. The two balls meet at t =18 s. What is the value of v? (take g = 10 m/s^2) (i) 75 m/s (ii) 55m/s (iii) 40m/s (iv) 60 m/s Please help bayya. correct answer is (i). Can you explain why? 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 a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with speed v. The two balls meet at t =18 s. What is the value of v? (take g = 10 m/s^2) (i) 75 m/s (ii) 55m/s (iii) 40m/s (iv) 60 m/s Please help bayya. correct answer is (i). Can you explain why? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with speed v. The two balls meet at t =18 s. What is the value of v? (take g = 10 m/s^2) (i) 75 m/s (ii) 55m/s (iii) 40m/s (iv) 60 m/s Please help bayya. correct answer is (i). Can you explain why?.
A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with speed v. The two balls meet at t =18 s. What is the value of v? (take g = 10 m/s^2) (i) 75 m/s (ii) 55m/s (iii) 40m/s (iv) 60 m/s Please help bayya. correct answer is (i). Can you explain why? 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 a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with speed v. The two balls meet at t =18 s. What is the value of v? (take g = 10 m/s^2) (i) 75 m/s (ii) 55m/s (iii) 40m/s (iv) 60 m/s Please help bayya. correct answer is (i). Can you explain why? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with speed v. The two balls meet at t =18 s. What is the value of v? (take g = 10 m/s^2) (i) 75 m/s (ii) 55m/s (iii) 40m/s (iv) 60 m/s Please help bayya. correct answer is (i). Can you explain why?.
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