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The position of a projectile launched fr...

The position of a projectile launched from the origin at t = 0 is given by `s`=`(40hati+50hatj)m`at `t=2s`. if the projectile was launched at an angle `theta` from the horizontal, then `theta`is (take g = 10 `ms^(–2)`

A

` tan ^( -1) "" (2)/(3) `

B

` tan ^(-1) "" (3)/(2) `

C

` tan ^(-1) "" (7)/(4) `

D

` tan ^(-1) "" (4)/(5) `

Text Solution

Verified by Experts

The correct Answer is:
B
`"as "vecr=40hati+50hatj`
At t=2sec `rArr u cos theta=x`
`u sin thetat-1/2 gt^(2)=y" Putting t=2"`
`rArr u cos theta=20 and u sin theta=35`
`rArr tan theta=7/4 rArr theta=tan^(-1) (7/4)`
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