A particle is projected from the ground with an initial speed of V at an angle of projection `theta`. The average velocity of the particle between its time of projection and time it reaches highest point of trajectory is

Options

A. (a) `V/(2)sqrt(1+2cos^(2)theta)`
B. (b) `V/(2)sqrt(1+2sin^(2)theta)`
C. (c ) `V/(2)sqrt(1+3cos^(2)theta)`
D. (d) `V costheta`

Answer

(c )

Explanation

Average Velocity = Displacement/Time
`V_(av) = sqrt(mu^(2) + R^(2)/4 )/(T/(2))` …..(1)
`mu = H_(max) = V^(2)sin^(2)theta/2g`,`R = V^(2)sin2theta/g`
Time of flight, `T = 2Vsintheta/g`
On substituting the values of `mu`,R ant T in equation (1)
`V_(av) = sqrt{(V^(2)sin^(2)theta/2g)^(2) + (V^(2)sin2theta/2g)^(2)}/(2Vsintheta/2g)`
`rArr V_(av)= (Vsintheta/2g)sqrt(V^(2)sin^(2)theta + 4V^(2)cos^(2)theta)/(Vsintheta/(g))`
`:. V_(av) = V/(2)sqrt(1 + 3cos^(2)theta)`