From a $20 m$ high tower one ball is thrown upward with speed of $10 m / s$ and another is thrown vertically downward at the same speed simultaneously. The time difference of their reaching the ground will be (in seconds) :

3

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2

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4

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Solution

The correct option is B $2$
Since in both cases, ball has same speed and thrown from same height of tower, so the time taken from tower to ground is same for both case because when first ball will reach the point of projection, it will also have same speed as was during projection. Thus, time difference will be the time of flight of first ball (upward thrown).
Thus, time of flight $= 2 u s i n 90 / g = 2 u / g = 2 (10) / 10 = 2 s e c$

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From a 20m high tower one ball is thrown upward with speed of 10m/s and another is thrown vertically downward at the same speed simultaneously. The time difference of their reaching the ground will be (in seconds) :

From a $20 m$ high tower one ball is thrown upward with speed of $10 m / s$ and another is thrown vertically downward at the same speed simultaneously. The time difference of their reaching the ground will be (in seconds) :