Recent content by VaitVhat52

Ah that's correct. So the final answer would be ##\sqrt{2gL(1-\frac{1}{2})}##

One last thing: Since it asks for the answer in terms of ##g## and ##L##, are we allowed to have ##Cosθ## as part of the answer?

No, it starts at the as high as the vertical component of L. Would that then make it √2g(L-LCosθ) or am I missing something?

Started by analyzing the change in energy from the initial position to the final position which gives us mgh=1/2mv^2 Since we are trying to find speed, we rearrange the equation to solve for v, which gives us √2gL. My question is, do we need to take a component of L for √2gL because it is at...