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A man is standing on the top of building 100 m high. He throws two balls vertically, One at t=0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is + 15 m at t=2 s. The gap is found to remain constant. Calculate the velocity with which balls were thrown and the exact time interval between their thrown.?
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Problem: A man throws two balls vertically from the top of a building 100 m high. One ball is thrown at t=0 and the other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is 15 m at t=2 s. The gap remains constant. Find the velocity with which the balls were thrown and the time interval between their throws.
Solution:
To solve this problem, we can use the equations of motion for vertical motion:
At t=0, the first ball is thrown with an initial velocity (u1y) and reaches a maximum height of 100 m. Using the position equation, we can find the initial velocity of the first ball:
y = u1yt - 1/2gt2
At the maximum height, the velocity of the first ball is zero. Therefore, we can substitute y=100 m and vy=0 into the equation:
100 m = u1y(t=2 s) - 1/2(9.8 m/s2)(2 s)2
Solving for u1y, we get:
u1y = 49 m/s
Therefore, the first ball was thrown vertically upwards with an initial velocity of 49 m/s.
The second ball is thrown with an initial velocity (u2y) that is half the initial velocity of the first ball (u1y/2). Using the same equation, we can find the initial velocity of the second ball:
y = u2yt - 1/2gt2
At t=2 s, the second ball is 15 m below the first ball. Therefore, we can substitute y=-15 m and u2y=u1y/2 into the equation:
-15 m = (u1y/2)(t=2 s) - 1/2(9.8 m/s2)(2 s)2
Solving for u2y,
Solution:
To solve this problem, we can use the equations of motion for vertical motion:
Equation of motion:
- Position: y = uyt - 1/2gt2
- Velocity: vy = uy - gt
- Acceleration: g = 9.8 m/s2
Step 1: Determine the initial velocity of the first ball (u1y)
At t=0, the first ball is thrown with an initial velocity (u1y) and reaches a maximum height of 100 m. Using the position equation, we can find the initial velocity of the first ball:
y = u1yt - 1/2gt2
At the maximum height, the velocity of the first ball is zero. Therefore, we can substitute y=100 m and vy=0 into the equation:
100 m = u1y(t=2 s) - 1/2(9.8 m/s2)(2 s)2
Solving for u1y, we get:
u1y = 49 m/s
Therefore, the first ball was thrown vertically upwards with an initial velocity of 49 m/s.
Step 2: Determine the initial velocity of the second ball (u2y)
The second ball is thrown with an initial velocity (u2y) that is half the initial velocity of the first ball (u1y/2). Using the same equation, we can find the initial velocity of the second ball:
y = u2yt - 1/2gt2
At t=2 s, the second ball is 15 m below the first ball. Therefore, we can substitute y=-15 m and u2y=u1y/2 into the equation:
-15 m = (u1y/2)(t=2 s) - 1/2(9.8 m/s2)(2 s)2
Solving for u2y,
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A man is standing on the top of building 100 m high. He throws two balls vertically, One at t=0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is + 15 m at t=2 s. The gap is found to remain constant. Calculate the velocity with which balls were thrown and the exact time interval between their thrown.?
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A man is standing on the top of building 100 m high. He throws two balls vertically, One at t=0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is + 15 m at t=2 s. The gap is found to remain constant. Calculate the velocity with which balls were thrown and the exact time interval between their thrown.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A man is standing on the top of building 100 m high. He throws two balls vertically, One at t=0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is + 15 m at t=2 s. The gap is found to remain constant. Calculate the velocity with which balls were thrown and the exact time interval between their thrown.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A man is standing on the top of building 100 m high. He throws two balls vertically, One at t=0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is + 15 m at t=2 s. The gap is found to remain constant. Calculate the velocity with which balls were thrown and the exact time interval between their thrown.?.
A man is standing on the top of building 100 m high. He throws two balls vertically, One at t=0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is + 15 m at t=2 s. The gap is found to remain constant. Calculate the velocity with which balls were thrown and the exact time interval between their thrown.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A man is standing on the top of building 100 m high. He throws two balls vertically, One at t=0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is + 15 m at t=2 s. The gap is found to remain constant. Calculate the velocity with which balls were thrown and the exact time interval between their thrown.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A man is standing on the top of building 100 m high. He throws two balls vertically, One at t=0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is + 15 m at t=2 s. The gap is found to remain constant. Calculate the velocity with which balls were thrown and the exact time interval between their thrown.?.
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