If the vertical initial speed of the ball is 2.0 as the cannon moves horizontally at a speed of 0.65 , how far from the launch point does the ball fall back into the cannon?

Question

anonymous · Answer

Remember to include units. Anyways, you use the kinematic equations for this. For displacement in a direction you have  $$y=y _{0}+v _{y}t+1/2 a_{y}t^2$$ plugging in you get $$0=0+2.0t-4.9t^2$$ you can solve for t. You will get two solutions but you can obviously throw out t = 0. Once you get the amount of time it was in the air all you need to do is multiply the horizontal velocity by the amount of time it was in the air and you get how far it went horizontally.

anonymous · Answer

How did you get y = to 0? Like why is it 0=0?

anonymous · Answer

Because you are saying it starts at its lowest position and you want to know how long it takes to go from the lowest position back to that position. Make sense?

anonymous · Answer

Not really. I don't understand why it would be 0=0

anonymous · Answer

When I finished the answer completely I also got a negative answer.

anonymous · Answer

y = displacement in y direction... 
displacement = change in position ..

change in position  = final position - initial position
since the cannon ball started out in the cannon .. and ends up back in the cannon.. the displacement = zero right?? cause initial and final positions are the same?

anonymous · Answer

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