Recent content by foonis

Thanks you helped a ton :D

so then i could just use distance=velocity x time so x=((g√h)/(√g))(2√(h/g)) so x=2gh/g so x=2h So the distance between the blocks after they land is 2h?

with that equation I got the velocity of Block A to be (g√h)/(√g) Am I any closer?

wait could I use the equation Vf = Vi + at ? im going to try that and see what i get

Wow this is a tough one haha hmm.. Well I used a kinematics equation to find the velocity as the block slides along the table after the other block hits the ground, which I set up like: X=Xo+VoT+.5at^2, which i substituded the values to get: X=0+(0)(2√(h/g))+.5(g/2)(2√(h/g))^2 so...

Oh, you're right, I didnt notice that and ran with it. So I plugged it all in again and I got b) 2h/g=t^2 as the base equation to get 2h/(g/2)=t^2 which should simplify to 2√(h/g)=t. e) I fixed it to end up with the time between the block(a) leaving the table and hitting the floor as...

you should use the equation y = yo +vit +.5gt^2 for vertical motion. also Vi should equal zero, as the boulder begins from rest in the y-axis and Vf should equal the final velocity as the boulder hits the ground

Homework Statement Two small blocks, each of mass m, are connected by a string of constant length (4h) and negligable mass. Block A is placed on a smooth tabletop as shown, and Block B hangs over the edge of the table. The tabletop is a distance (2h) above the floor. Block B is then...