Recent content by deanine3

Your right, I just get so overwhelmed. The time is squared so it would be the square root of 15.1= 3.88s. I appreciate your help. I am just stuck, I don't know how to solve this equation. I just am not sure what information I need, and then what to do with it. I DO see that the height, mass and...

h(20)= 1/2*9.8m/s*t^2 Is time 15.1 s?

Here is what I thought if her Vi=19.79 m/s (not sure of units) then to solve for b, KE=0.5*60*19.79= 593.7. Right?

I am not sure how to get the answer for (a). I though I was on the right track. The gpe on the balcony is 11,760, the KE on the balcony is 0. Then, she takes off... Her GPE decreases and her KE increases until she hits the water. Once she hits the water her GPE is 0 and KE is 11,760. Except to...

Is the answer 19.79 m/s the answer for a? or am I solving for nothing? haha

60*9.8*20=1/2*60*v^2 11,760 = 30 v^2 both sides /30 v=square root of 392 = 19.79 ?

oh! Sorry. I'm new at physics ;) to solve for t y = Vi(t) + g(t^2) and solve for t. Since she is starting from rest (assumably) Vi is 0(t) + 9.8(t^2), then...I'm not sure. Could you give me a nudge.

Is it ok to "borrow" that info from the second part of the question? I don't know the rules of physics.

Where is the 60kg from? is it 20m*3m?

Homework Statement Judy looks from her 20m high balcony to a swimming pool below, not directly below, but 3m out from the bottom of the building. If the pool is 6m wide. (a) How fast would she have to jump horizontally to land in the middle of the pool? (b) Judy's mass is 60 kg, what will be...

oh, sorry, h= (1/2)(10 m/s^2)/(9.8 m/s^2)= 5.10? Does that look right? Am I supposed to divide 0.5 x 10 x 10 then/ 9.8 or is it 0.5 x 10 (don't square)/9.8? It seems like the 10 is squared and the 9.8 has the seconds squared.

Homework Statement Suppose a car starts coasting from the top of a hill that is 60m high. (a)How fast will it be going at the bottom of the hill if there is no friction? (b) For the same car starting at the top of another hill and reaching the bottom, without friction, at 10 m/s, how high is...

Ok, I really think I have it! When you divide by 2, it becomes 2gh=v^2. The v^2 becomes v=square root sign. Then, under the square root sign you plug in the 2 (from the previous division), gravity, and the height! This is slowly making sense.

or are both sides multiplied by 2?

I think I am getting however, how do you take out the 1/2 from the v^2 side of the equation? I don't see where it went.