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Standard XII

Physics

1st Equation of Motion

A ball is thr...

Question

A ball is thrown vertically upwards with a speed of $10 m / s$ from the top of a tower $200 m$ height and another is thrown vertically downwards with the same speed simultaneously. The time difference between them on reaching the ground is $(g = 10 m / s^{2})$

12 s

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6 s

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2 s

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1 s

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Solution

The correct option is D $2 s$
$1^{s t}$ ball will return to initial pt. with a velocity which is same in magnitude and reverse in direction, hence from A to C the time is same.
So, difference is when ball from A goes to B
and comes back to A.
Hence, this time, $t = \frac{2 u}{g} = \frac{2 \times 10}{10} = 2 s e c$

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A ball is thrown vertically upwards with a speed of 10 m/s from the top of a tower 200 m height and another is thrown vertically downwards with the same speed simultaneously. The time difference between them on reaching the ground is (g=10 m/s2)

The correct option is D 2 s1st ball will return to initial pt. with a velocity which is same in magnitude and reverse in direction, hence from A to C the time is same.So, difference is when ball from A goes to Band comes back to A.Hence, this time, t=2ug=2×1010=2 sec

A ball is thrown vertically upwards with a speed of $10 m / s$ from the top of a tower $200 m$ height and another is thrown vertically downwards with the same speed simultaneously. The time difference between them on reaching the ground is $(g = 10 m / s^{2})$