Recent content by kat R

That works! Thank you so much!

It points in the -y direction... so the final equation would be h=d*tanθ-.5*g*(d/Vo*cosθ)^2

So it would be h=d*tanθ+.5*g*(d/Vo*cosθ)^2

So I replaced the Yf and Yi for h and replaced Vyi and got h=Vo*sin(θ)*t+.5*ay*t^2 Then the final equation I got was h=Vo*sin(θ)*[d/(Vo*cos(θ)]+.5*ay*[d/Vo*cos(θ)]^2

So if the equation is d=Vo*cos(θ)*t I can say that t=d/(Vo*cos(θ)) but I can't go any further because I wasn't given any numerical values

Okay so ax would be 0 which would leave the equation at d=Vxi*t

Yea Xf-Xi would equal d since Xi is 0. Depends if acceleration due to gravity would matter in the x-direction

I think Xi could be taken out but I don't think anything else can

Xf=Xi+Vxi*t+.5*ax*t^2

Homework Statement (For this problem ignore the height of the firefighter) A firefighter, a distance (d) from a building, shoots a stream of water at an initial angle, above the horizontal, at initial speed (Vo). At what height (h) does the water strike the building? No values were given to...

I'm a new student taking Engineering Physics.