yaah sohil is right.........

u may also approach like

if the 2 pts (A,B)on the trajectory lie at equal distance from max height "(on opposite sides-------as max height is b/w them) then

the vertical displacement from pt A to B is 0

thus Vy avg=0/t=0

but the horizontal displacement is ucosq *t in time t

thus

Vx avg=ucosq/t *t=ucosq

whr q=angle made by velocity vector with the horizontal axis

thus total avg=ucosq+0

hopr u got it.....if still confused nudge me

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