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)

Question

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 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

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

`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`

5 views Answered 2023-02-05 15:29