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A particle is projected with a velocity ...

A particle is projected with a velocity of 50 m/s at `37^@` with horizontal. Find velocity, displacement and co-ordinates of the particle (w.r.t. the starting point) after 2 s.
Given, `g=10m//s^2, sin 37^@ = 0.6 and cos 37^@ = 0.8`.

A

`(40hati + 10hatj) m//s` `x=80 m and y= 80m`

B

`(40hati + 10hatj) m//s` `x=20 m and y= 40m`

C

`(40hati + 10hatj) m//s` `x=80 m and y= 40m`

D

`(40hati + 20hatj) m//s` `x=80 m and y= 40m`

Text Solution

Verified by Experts

The correct Answer is:
C
In the given problem,
` u=(50cos 37^@)hat(i) + (50sin37^@)hatj`
`=(40hati + 30hatj)m//s`
`a=(-10hatj)m//s^2`
` t=2s`
`v=u+at`
`=(40hati +30 hatj)+(-10hatj)(2)`
`=(40hati + 10hatj) m//s`

`s=ut+1/2at^2`
`=(40hati + 30hatj)(2)+ 1/2(-10hatj)(2)^2`
`=(80hati + 40hatj)m`
Coordinates of the particle are
x=80 m and y= 40m .
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