Find the average velocilty of a projectile between the instants it crosses half the maximum height. It is projected with speed u ast angle `theta` with the horizontal.

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1 Answers

`u cos theta` along horizontal
avg. velocity is a vector
first we will find vertical component
`V_(y)=("Total Displacement")/("Total time")=0`
Horizontal
`V_(1 x)=V_(2 x)=u cos theta`
`V_(x)=u cos theta`

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