A small 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
