Question

An orange is tossed upward at 21m/s. What is the velocity of the orange 3.5s later. What is the height of the orange at this time. Is the orange still traveling up or is it traveling down.

Answer 1

a)After 3,5s -->v=v0+gt=21+(-9,8•3,5)=21+(-34,3)=-13,3m/s; b)The maximum height that the orange reaches is h max=v0^2/2g=22,5m;v^2=sqrt(2gh)=>h=v^2/2g=9,025m.The height of the orange is H=h max-h=13,475m.; c)The orange is traveling down.