Recent content by Baiatul122001

But the efficiency of the circuit does not depend on the electromotive voltage, is the answer wrong?

The answer in the textbook's

Maybe the textbook's solution is wrong

That's what you have to calculate.

Let ##d## be the distance between A and B when B hits the ground. For a given ##v_1## find ##v_2## such that ##d## is minimised. What is the expression of "d" before minimization?

Check how you got the h/g at the end of this: ##H=v_1t-gt^2/2=v_1√(2h/g)-h/g##. I used the time obtained for the body from B to touch the ground and using the law of uniform rectilinear motion:H=v1t-gt^2/2=v1√(2h/g)-g[√(2h/g)]^2/2=v1√(2h/g)-g(2h/g)/2=v1√(2h/g)-h Ahh, I was wrong the first time

So it's not correct with functions, ahhh

##v_1√(2h/g)-h/g## cannot be right. The left hand term is a distance, the right hand term is time2. Similarly I see terms (hg-h) later, which make no sense. You should always check for dimensional consistency.It is squared because L ^ 2 = l ^ 2-2l√ (2h / g) v2 + (2hv2 ^ 2 + 2hv1 ^ 2) / g +...

For v1 >> 0, the point where they are closest is at t = 0, ie when the bodies are in the initial position have the minimum distance between them?

During t the body from A climbs: H=v1t-gt^2/2=v1√(2h/g)-h/g. So the height at which the body in A when the one in B reaches the ground is H'=H+h=v1√(2h/g)-h/g+h=v1√(2h/g)+(-h+hg)/g. The horizontal distance between the bodies at this time is D=l-d=l-v2√(2h/g). Now we apply the Pythagorean Theorem...

We have to calculate the speed v2 so that the distance between the two is minimal. The answer must be:

I do not know if he has reached the maximum height or has continued to reach the maximum height

It's in the figure next to where I got stuck