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A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 22 seconds). The later ball is thrown at a velocity of half the first. At t = 2 s, both the balls reach their maximum heights. At this time the vertical gap between first and second ball is +15m.
The speed of the first ball is
  • a)
    20 m s−1
  • b)
    10 m s−1
  • c)
    5 m s−1
  • d)
    15 m s−1
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A man is standing on top of a building 100 m high. He throws two ball...
Given information:
- A man is standing on top of a building 100 m high.
- Two balls are thrown vertically.
- The second ball is thrown after a time interval (less than 22 seconds) with a velocity half of the first ball.
- At t = 2 s, both balls reach their maximum heights.
- At this time, the vertical gap between the first and second ball is 15 m.

To find:
The speed of the first ball.

Explanation:

Step 1: Understanding the problem
- We have a man on top of a building who throws two balls vertically.
- The second ball is thrown later and with half the velocity of the first ball.
- We need to find the speed of the first ball.

Step 2: Analyzing the situation
- Let's assume the initial velocity of the first ball is V m/s.
- The initial velocity of the second ball will be V/2 m/s.
- Both balls reach their maximum heights at t = 2 s.

Step 3: Calculating the maximum height
- The maximum height reached by a ball thrown vertically can be calculated using the formula: H = (V^2)/(2g)
- For the first ball, the maximum height will be H1 = (V^2)/(2g).
- For the second ball, the maximum height will be H2 = ((V/2)^2)/(2g) = (V^2)/(8g).
- Given that the vertical gap between the two balls at t = 2 s is 15 m, we can write the equation: H1 - H2 = 15.

Step 4: Solving the equations
- Substituting the values of H1 and H2 in the equation, we get: (V^2)/(2g) - (V^2)/(8g) = 15.
- Simplifying the equation, we have: (4V^2 - V^2)/(8g) = 15.
- Further simplifying, we get: (3V^2)/(8g) = 15.
- Multiplying both sides by (8g/3), we have: V^2 = 15 * (8g/3).
- Taking the square root of both sides, we get: V = sqrt(15 * (8g/3)).

Step 5: Calculating the speed of the first ball
- The speed of the first ball is given by the initial velocity, which is V.
- Substituting the value of V, we have: speed of first ball = sqrt(15 * (8g/3)).
- Since the acceleration due to gravity (g) is approximately 10 m/s^2, we can calculate the speed of the first ball as:
- speed of first ball = sqrt(15 * (8 * 10/3)) = sqrt(400) = 20 m/s.

Therefore, the speed of the first ball is 20 m/s, which corresponds to option 'A'.
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Community Answer
A man is standing on top of a building 100 m high. He throws two ball...
Let the speeds of the two balls, 1 and 2, be v1 and v2, respectively. Since the speed of the second ball is half of the first, let
v1 = 2v ...(1)
v2 = v ...(2)
If y1 and y2 are the maximum heights reached by the balls 1 and 2, respectively, then
Since it is given that y1 − y2= 15 m
Substituting in equation
v1 = 20 m s−1
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A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 22 seconds). The later ball is thrown at a velocity of half the first. At t = 2 s, both the balls reach their maximum heights. At this time the vertical gap between first and second ball is +15m.The speed of the first ball isa)20 m s−1b)10 m s−1c)5 m s−1d)15 m s−1Correct answer is option 'A'. Can you explain this answer?
Question Description
A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 22 seconds). The later ball is thrown at a velocity of half the first. At t = 2 s, both the balls reach their maximum heights. At this time the vertical gap between first and second ball is +15m.The speed of the first ball isa)20 m s−1b)10 m s−1c)5 m s−1d)15 m s−1Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 22 seconds). The later ball is thrown at a velocity of half the first. At t = 2 s, both the balls reach their maximum heights. At this time the vertical gap between first and second ball is +15m.The speed of the first ball isa)20 m s−1b)10 m s−1c)5 m s−1d)15 m s−1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 22 seconds). The later ball is thrown at a velocity of half the first. At t = 2 s, both the balls reach their maximum heights. At this time the vertical gap between first and second ball is +15m.The speed of the first ball isa)20 m s−1b)10 m s−1c)5 m s−1d)15 m s−1Correct answer is option 'A'. Can you explain this answer?.
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