Resolução do Desafio de Mestre de 16 / 07

Lançamentos oblíquos

Pedra 1:

y = v0 . senθ. t + gt²/2   = >   2,0 = 3,0 . t + 5,0 . t²   = >   t = 0,40s
x1 = v0 . cosθ . t   = >   x1 = 4,0 . 0,40(m) = 1,6 m

Pedra 2:

y = - v0 . senθ. t + gt²/2   = >   2,0 = - 3,0 . t + 5,0 . t²   = >   t = 1,0s
x2 = v0 . cosθ . t   = >   x2 = 4,0 . 1,0(m) = 4,0 m

Δt = 1,0s - 0,40s = 0,60s

Resposta: A