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A girl standing on a stationary lift (open from above) throws a ball upwards with initial speed 50 m s-1. The time taken by the ball to return to her hands is (Take g = 10 ms-2)
- a)5s
- b)10 s
- c)15 s
- d)20 S
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A girl standing on a stationary lift (open from above) throws a ball u...
Since the ball reaches back so distance will be zero :. s =0
using second equation of motion
s=ut +1/2 at²
0= 50×t + 1/2 (-10t²)
0= 50t -5t²
50 t = 5t²
50 = 5t
t = 10 sec :)
using second equation of motion
s=ut +1/2 at²
0= 50×t + 1/2 (-10t²)
0= 50t -5t²
50 t = 5t²
50 = 5t
t = 10 sec :)
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Community Answer
A girl standing on a stationary lift (open from above) throws a ball u...
Analysis:
The ball is thrown upwards from a stationary lift with an initial speed of 50 m/s. The acceleration due to gravity is 10 m/s^2. We need to find the time taken for the ball to return to the girl's hands.
Calculations:
- The time taken for the ball to reach its maximum height can be calculated using the formula:
t = (Vf - Vi) / g
t = (0 - 50) / -10
t = 5 s
- The time taken for the ball to return to the girl's hands is twice the time taken to reach the maximum height, as the motion is symmetrical.
Therefore, the total time taken is 2 * 5 = 10 s.
Therefore, the correct answer is option B) 10 s.
The ball is thrown upwards from a stationary lift with an initial speed of 50 m/s. The acceleration due to gravity is 10 m/s^2. We need to find the time taken for the ball to return to the girl's hands.
Calculations:
- The time taken for the ball to reach its maximum height can be calculated using the formula:
t = (Vf - Vi) / g
t = (0 - 50) / -10
t = 5 s
- The time taken for the ball to return to the girl's hands is twice the time taken to reach the maximum height, as the motion is symmetrical.
Therefore, the total time taken is 2 * 5 = 10 s.
Therefore, the correct answer is option B) 10 s.
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A girl standing on a stationary lift (open from above) throws a ball upwards with initial speed 50 m s-1. The time taken by the ball to return to her hands is (Take g = 10 ms-2)a)5sb)10 sc)15 sd)20 SCorrect answer is option 'B'. Can you explain this answer?
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