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## Homework Statement

A cannon is firing from an elevated position 100 m above a target 800 m away. The ititial velocity of the cannon is 500 m/s.

## Homework Equations

Vx = VCOSq

Vy = VSINq

a = (Vf - Vi)/(t)

Vf = Vi + a t

s = 0.5 (Vo +Vf)t

Vf2 = Vo2 + 2aS

S = Vit + 0.5at2

## The Attempt at a Solution

What I tried was dividing the problem in two pieces and working backwards. The angle of reach to get to a position level with the cannon is theta = 0.5(arcsin (gx/v^2)) So if I can get the distance from the target where the cannonball is level with cannon I can go back and get the original angle. At that level point the velocity should be the same as when it left the cannon but going down.

The time to get to the target should be the distance in the x direction from that point Xb times Vx = t = Xb/(VxCOS q). The time of flight should be found out by calculating s = -VSINq t + (0.5)at^2 and the other kinematics equations. But I keep going around in cirles on this.