Asmall doll is thrown from a window with an initial speed of 6 m/s at 35 degrees above the horizontal. what is its velocity of the doll one second later

It will be 9.55 m/s.
as the velocity vector will be
6.38 in vertical direction pointing downwards after 1 second
and the horizontal will be 4.86
the angle between them is 35°
thus using parallelogram law of vector addition
you get
v^2 = 6.38*6.38+4.86*4.86+2(6.38)(4.86)cos35°
v^2 = 40.7044+23.61+26.92
v^2 = 91.24
v = 9.55 m/s

By dividing you can find the answer