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A stone is dropped by a person from the top of building , which is 200m tall . At the same time , another stone is thrown upwards , with a velocity of 50 m/s by a person standing at the foot of the building . Find the time after which the stones meet ? Answer is 4 seconds ! Please give me solution .?
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A stone is dropped by a person from the top of building , which is 200...
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A stone is dropped by a person from the top of building , which is 200...
Solution:
We can use the formula of motion under gravity to solve this problem.
For the stone dropped from the top of the building:
Initial velocity u = 0
Acceleration due to gravity a = 9.8 m/s^2
Distance s = 200 m
We can use the formula s = ut + 1/2 at^2 to find the time taken by the stone to reach the ground.
200 = 0*t + 1/2*9.8*t^2
t^2 = 200/4.9
t = sqrt(40.8)
t = 6.4 seconds
Therefore, the stone takes 6.4 seconds to reach the ground.
For the stone thrown upwards:
Initial velocity u = 50 m/s (upwards)
Acceleration due to gravity a = 9.8 m/s^2
Distance s = 200 m
We can use the formula s = ut + 1/2 at^2 to find the time taken by the stone to reach the highest point.
200 = 50*t + 1/2*(-9.8)*t^2
4.9t^2 - 50t + 200 = 0
We can solve this quadratic equation to find the time taken by the stone to reach the highest point.
Using the quadratic formula, we get:
t = (50 ± sqrt(50^2 - 4*4.9*200))/(2*4.9)
t = (50 ± sqrt(2500 - 3920))/9.8
t = (50 ± sqrt(-1420))/9.8
Since the square root of a negative number is not a real number, the stone will never reach the highest point. Instead, it will reach a certain height and then start falling back down.
Let's find the height reached by the stone before it starts falling back down.
Using the formula v^2 = u^2 + 2as, we get:
0 = 50^2 - 2*9.8*s
s = 127.55 m (approx)
Therefore, the stone reaches a height of approximately 127.55 m before it starts falling back down.
Now, we can use the formula s = ut + 1/2 at^2 to find the time taken by the stone to reach the ground.
127.55 = 50*t + 1/2*9.8*t^2
4.9t^2 + 50t - 127.55 = 0
We can solve this quadratic equation to find the time taken by the stone to reach the ground.
Using the quadratic formula, we get:
t = (-50 ± sqrt(50^2 - 4*4.9*(-127.55)))/(2*4.9)
t = (-50 ± sqrt(10000 + 2491.1))/9.8
t = (-50 ± sqrt(12491.1))/9.8
Since the time taken by the stone cannot be negative, we take the positive value of t.
t = (-50 + sqrt(12491.1))/9.8
t = 4 seconds (approx)
Therefore, the stones will meet after 4 seconds.
We can use the formula of motion under gravity to solve this problem.
For the stone dropped from the top of the building:
Initial velocity u = 0
Acceleration due to gravity a = 9.8 m/s^2
Distance s = 200 m
We can use the formula s = ut + 1/2 at^2 to find the time taken by the stone to reach the ground.
200 = 0*t + 1/2*9.8*t^2
t^2 = 200/4.9
t = sqrt(40.8)
t = 6.4 seconds
Therefore, the stone takes 6.4 seconds to reach the ground.
For the stone thrown upwards:
Initial velocity u = 50 m/s (upwards)
Acceleration due to gravity a = 9.8 m/s^2
Distance s = 200 m
We can use the formula s = ut + 1/2 at^2 to find the time taken by the stone to reach the highest point.
200 = 50*t + 1/2*(-9.8)*t^2
4.9t^2 - 50t + 200 = 0
We can solve this quadratic equation to find the time taken by the stone to reach the highest point.
Using the quadratic formula, we get:
t = (50 ± sqrt(50^2 - 4*4.9*200))/(2*4.9)
t = (50 ± sqrt(2500 - 3920))/9.8
t = (50 ± sqrt(-1420))/9.8
Since the square root of a negative number is not a real number, the stone will never reach the highest point. Instead, it will reach a certain height and then start falling back down.
Let's find the height reached by the stone before it starts falling back down.
Using the formula v^2 = u^2 + 2as, we get:
0 = 50^2 - 2*9.8*s
s = 127.55 m (approx)
Therefore, the stone reaches a height of approximately 127.55 m before it starts falling back down.
Now, we can use the formula s = ut + 1/2 at^2 to find the time taken by the stone to reach the ground.
127.55 = 50*t + 1/2*9.8*t^2
4.9t^2 + 50t - 127.55 = 0
We can solve this quadratic equation to find the time taken by the stone to reach the ground.
Using the quadratic formula, we get:
t = (-50 ± sqrt(50^2 - 4*4.9*(-127.55)))/(2*4.9)
t = (-50 ± sqrt(10000 + 2491.1))/9.8
t = (-50 ± sqrt(12491.1))/9.8
Since the time taken by the stone cannot be negative, we take the positive value of t.
t = (-50 + sqrt(12491.1))/9.8
t = 4 seconds (approx)
Therefore, the stones will meet after 4 seconds.
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A stone is dropped by a person from the top of building , which is 200m tall . At the same time , another stone is thrown upwards , with a velocity of 50 m/s by a person standing at the foot of the building . Find the time after which the stones meet ? Answer is 4 seconds ! Please give me solution .?
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A stone is dropped by a person from the top of building , which is 200m tall . At the same time , another stone is thrown upwards , with a velocity of 50 m/s by a person standing at the foot of the building . Find the time after which the stones meet ? Answer is 4 seconds ! Please give me solution .? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A stone is dropped by a person from the top of building , which is 200m tall . At the same time , another stone is thrown upwards , with a velocity of 50 m/s by a person standing at the foot of the building . Find the time after which the stones meet ? Answer is 4 seconds ! Please give me solution .? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone is dropped by a person from the top of building , which is 200m tall . At the same time , another stone is thrown upwards , with a velocity of 50 m/s by a person standing at the foot of the building . Find the time after which the stones meet ? Answer is 4 seconds ! Please give me solution .?.
A stone is dropped by a person from the top of building , which is 200m tall . At the same time , another stone is thrown upwards , with a velocity of 50 m/s by a person standing at the foot of the building . Find the time after which the stones meet ? Answer is 4 seconds ! Please give me solution .? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A stone is dropped by a person from the top of building , which is 200m tall . At the same time , another stone is thrown upwards , with a velocity of 50 m/s by a person standing at the foot of the building . Find the time after which the stones meet ? Answer is 4 seconds ! Please give me solution .? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone is dropped by a person from the top of building , which is 200m tall . At the same time , another stone is thrown upwards , with a velocity of 50 m/s by a person standing at the foot of the building . Find the time after which the stones meet ? Answer is 4 seconds ! Please give me solution .?.
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