A particle is projected from the ground with an initial speed of v at an angle `theta` with horizontal. The average velocity of the particle between its point of projection and highest point of trajectroy is :

Options

A. `u cos theta`
B. `u/2 sqrt(1+cos^(2) theta)`
C. `u/2 sqrt(1+2cos^(2) theta)`
D. `u/2 sqrt(1+3cos^(2) theta)`

Answer

D

Explanation

From figure,
Average velocity = `("displacement ")/("time taken")`
`:.v_(av)=(sqrt(H^2+R^2/4))/(T//2)`
Hence , `H =(u^2sin^2 theta)/(2g) , R=(u^2sin 2 theta)/g and T=(2usintheta)/g`
`:.v_(av)=sqrt(((u^2sin^2theta)/(2g))^2+1/4 ((u^2sin2 theta)/g)^2)/(((2usintheta)/(2g)))`
` =sqrt(((u^2sin^2theta)/(2g))^2+ ((2u^2sinthetacos theta)/g)^2)/((usintheta)/(g))`
` =sqrt(((u^2sin^2theta)/(2g))(sin^(2) theta+4cos^2theta))/((usintheta)/g)=u/2sqrt(1+3 cos^2theta)`