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A particle is projected in the X-Y plane...

A particle is projected in the `X-Y` plane. `2sec` after projection the velocity of the aprticle makes an angle `45%(@)` with the `X-`axis. `4sec` after projection, it moves horizontally. Find the velocity of projection (use `g = 10 ms^(-2))`.

Text Solution

Verified by Experts

The correct Answer is:
`20sqrt(5)m//s`
`V_(y) = u_(y) +a_(y)t`
at heighest point of motion, `V_(y) = 0`
`0 = u_(y) - g xx 4`
`u_(y) = 4g`
`u_(y) = 40m//s`

at `t = 2 sec`,
`V_(y) = u_(y) +a_(y) t`
`V_(y) = 4g - gt rArr V_(y) = 20m//s`
given that `V_(y) = u_(x) = 20m//s`
velocity of projection `u = sqrt(u_(x)^(2) +u_(y)^(2))`
`rArr u = sqrt((40)^(2) +(20)^(2))`
`u = sqrt(2000) rArr u = 20 sqrt(5)m//s`
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