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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. 1.8mB. 0.9mC. 0.7mD. 0.11m |
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Answer» `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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