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A particle is projected fro the ground with an initial speed v at an angle theta withhorizontal. The average velocity of the particle between its point of projection and highest point trajectory is |
Answer» Solution : `vecv_(AVG)=(vecv+vecu)/(2)=(U cos THETA hati)+(u cos theta hati+ u SIN theta HATJ))/(2)` `v_(av)=(v)/(2) sqrt(1+3 cos^(2) theta)` |
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