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A particle is projected at an anlge tan^...

A particle is projected at an anlge `tan^(-1)((1)/sqrt(3))`. Find the height at which velocity of the particle is `9hati+3hatj` (m/sec). `(g=10ms^(-2))`

A

1.8m

B

0.9m

C

0.7m

D

0.11m

Text Solution

Verified by Experts

`tan theta=(u_(y))/(u_(x)), (1)/sqrt(3)=(u_(y))/(u_(x))=(u_(y))/(9)`
`u_(y)=3sqrt(3)`
`V_(y)^(2)=u_(x)^(2)+2a_(y)s_(y)`
`9=27+2(-10)S_(y)`
`S_(y)=0.9`
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