Recent content by esmikell

The velocity vector for the projectile on a moving railcar, truck, whatever is going to be v = [5+vcos(theta)]i + [vsin(theta)]j The 5m/s is added to the i component (x-dir). You can't go and add 5m/s to the y-dir.

Hello HallsofIvy, The cannon is on the truck. The truck is moving at 5m/s, let's say moving to the east. The cannon is fired with an unknown angle. The projectile has a velocity of 50 m/s at the unknown angle. The cannon is fired in the east direction.

Max range is always at 45 when you projectile is launched from an object that is not moving in the x-direction. If you fire a cannon that is on a railcar moving with a velocity, a 45 degree cannon will not give a max range. Since there is already a x-velocity you can then steal some...

Good morning, Thanks for looking into it. I will see what I can do with the cos's. Stephen

The triangles method uses a 45-45-90. From the top angle drop a line down to the base leg. The left side is 5 and the other is x. The side in common is h, the maximum height. Since this is a 45-45-90, the 5+x side is equal to the h. There is enough variables equal to each other that the angle in...

Hello everyone, I have a physics question for you all. A cannon fired from a truck at 5m/s. A cannon is fired in the same direction as the cannon. The velocity of the projectile is 50m/s. What angle will provide the greatest (distance in the x-driection) range for the projectile. I have...