A person in an elevator accelerating upwards with an acceleration of 2m/s2, tosses a coin vertically upwards with a speed of 20m/s. After how much time will the coin fall back into his hand? (Take g= 10m/s2)
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Solution :
As per the given data ,
- Acceleration of the elevator = 2 m/s ²
- Acceleration due to gravity = 10 m/s²
Net acceleration of the coin (a)= 10 + 2 = - 12 m/s ( as the coin is moving against gravity )
- Initial velocity of the coin = 20 m/s
- Final velocity ( at the highest point ) = 0 m/s
As the coin is moving with uniform acceleration throughout it's motion we can use the first equation of motion in order to find time .
- v = u + at
Now let's substitute the given values ,
⇒ 0 = 20 - 12 x t
⇒ t = 20 / 12
⇒ t = 1.6 s (approx.)
We know that ,
- Time of ascent = Time of descent = 1.6s
Hence ,
The total time taken by the ball to fall back into hand of the person
= 1.6 + 1.6
= 3.2 s
The total time taken by the ball to fall back into hand of the person is 3.2 s .
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