A stone is thrown from the edge of the top of a 40 m tall building, at some unknown angle,θ, above the horizontal. The stone strikes the ground a horizontal distance of 45 m from the base of the building, 5.0 s after beingthrown.Assume that the ground is level and that the side of the building is vertical. Determine the initial speed, vo with which the stone was thrown. one of these:Vo = 30.2 m/s Vo = 28.4 m/s Vo = 24.5 m/s Vo = 18.4 m/s Vo = 12.4 m/s Vo = 10.5 m/s
please show work if you can
the height y, is
y = h + vsinθ t - 4.9t^2
since y(5) = 0,
40 + 5v sinθ - 122.5 = 0
5v cosθ = 45
solve that to get v = 18.80 m/s
Not sure how they came up with 18.4