Recent content by bmore007

n/m :frown:

yes that is the starting point except it's (10/ViCos30)^2 the whole thing squared

unfortunately I'm not too good at more involved algebra like this :frown: (1/2)(-9.8)(10/ (ViCos30) )^2 -4.9(100/Vi^2* .75) so 4.4Vi^2 -4.9(100/.75Vi^2) :frown:

anychance you can give me a walk through numerically to the answer? I'm afraid that'll help me more then me struggling here

i don't understand why that isn't correct ViSin(30)*10 * ViCos(30) Vi.5 * 10 * Vi .87 Vi 4.4

"Can you factor "(ViSin30)10 * ViCos30" ?" 4.33Vi

all the #'s I'm getting are turning out wrong

1.05 = (ViSin30)10 * ViCos30+ 1/2(-9.8)(10/ViCos30)^2 help me solve for Vi please

1. get x1 in terms of t1 from the equation of motion, where x1 is the distance the rock travels from the top of the well to the water. Actually, mass doesn't enter into the equation xf1 = xi1 + Vox1 (t1) xf1-xi1 = Vox1 (t1) xf1-xi1/Vox1 = t1 2. get x2 in terms of t2 from the...

no mass for the rock, everything that is given is in the problem above

this program is saying my answer is incorrect

here's what i know Yi= h Yf= 0 t= 2.3s Vo= 336m/s g = -9.8 and i want to use Yf-Yi= Vot + 1/2(g)^2 correct? 1/2g^2 cancels out because of there being no g -h = 336(2.3) i guess my way of using the formula is wrong?

yf-yi = Vot ?? man I'm stupid

"1a is straightforward. You are given time and asked to find the distance. You should be able to write an equation of motion that relates distance, velocity, and time. Right? You know 2 of the 3 parameters, so solve for distance. Hint: think of the splash sound as a particle that starts moving...

thank you i got #2 correct, I'm still stuck on the first one though how would i go about finding the time it would take you to hear the splash for a well of depth h?