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A particle is projected upwards with vel...

A particle is projected upwards with velocity `20m//s`. Simultaneously another particle is projected with velocity `20(sqrt2) m//s` at `45^@`. `(g = 10m//s^2)`
(a) What is acceleration of first particle relative to the second?
(b) What is initial velocity of first particle relative to the other?
(c) What is distance between two particles after 2 s?

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
(a)`a_1 = a_2 = g` (downwards)
`:.` Relative acceleration = 0
(b)`u_(12) = u_1 -u_2`
`=(20hatj)-(20hati + 20hatj)`
`=(-20hati) m//s`
or `20m//s` in horizontal direction.
(c) `u_(12) = (-20hati) m//s` is constant.
Therefore relative motion is uniform.
`:. d = |u_(12)|t`
`=20 xx 2 = 40m` .
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