Recent content by Cassomophone

Could I do this instead? Since the equation is at 45 ° doesn't (Vx)i = (Vy)i? So if that is true... then equation 1 gives: 15000m=(Vx)i Δt equation 3 gives: 0=300+(Vy)iΔt-(1/2)(g)(Δt)^2 but since (Vx)i = (Vy)i then (Vy)iΔt = (Vx)i Δt = 15 so: 0=300+15000-(1/2)(g)(Δt)^2 so...

Ok so look at this real quick. Since the equation is at 45 ° doesn't (Vx)i = (Vy)i? So if that is true... then equation 1 gives: 15000m=(Vx)i Δt equation 3 gives: 0=300+(Vy)iΔt-(1/2)(g)(Δt)^2 but since (Vx)i = (Vy)i then (Vy)iΔt = (Vx)i Δt = 15 so: 0=300+15000-(1/2)(g)(Δt)^2 so...

Ok... I get that I can, but how do I do that then?

So I isolate the time from eq 1, but I still don't know the initial velocity of x, so how do I do that?

Ah true, I didn't mean 3 km, I meant to type .3km. But you are right I should convert the km into meters so, 15km=15,000m and 5km is 5,000 m. Thank you so much for looking at this.

Homework Statement So here is the question my professor made up. I asked if the time was missing and once he thought about it, he said we should be able to figure it out. I need help. A cannon is mounted on a railroad track on a cliff above the ocean. 300 m below, and standing...