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A man tosses object with initial velocity of 20 m/s into a room, and is 2.0 m above the ground. The room is 10.0 m above the ground. In a provided diagram, the distance from man to room horizontally is 31.8 m. I am required to solve for the angle of the object as it leaves the mans hand. vix= 20cosθ viy= 20sinθ To solve for θ, I must use the initial velocity given, broken up into components. Since I am given the vertical displacement (8.0 m) and horizontal displacement (31.8 m) , I can use 2 formulas and substitute one into the other, by isolating t as it is common to both. dx= vixt 31.8 = 20cosθt t= 31.8 / 20cosθ dy= viyt + 0.5ayt2 8.0 = 20sinθt - 4.9t2 8.0 = 20sinθ(31.8 / 20cosθ) - 4.9(31.8 / 20cosθ)2 8.0 = (31.8sinθ / cosθ) - 4.9(2.5281 / cos2θ) 8.0 = (31.8sinθ / cosθ) - (12.39 / cos2θ) and now i'm stuck.... I tried moving to one side and changing to common denominator... 8.0 = (31.8sinθ / cosθ) - (12.39 / cos2θ) 0 = (31.8sinθ / cosθ) - (12.39 / cos2θ) - 8.0 0 = (31.8sinθcosθ - 12.39 - 8.0cos2θ) / cos2θ 0 = 31.8sinθcosθ - 12.39 - 8.0cos2θ 0 = 15.9sin2θ - 12.39 - 8.0(1 - sin2θ) 0 = 15.9sin2θ - 12.39 - 8.0 + 8sin2θ 0 = 8sin2θ + 15.9sin2θ - 20.39 but no luck... i thought it looked like a quadratic at first, but I don't think it works with sin2θ any suggestions? thank you

Hello, I'm looking to apply to engineering programs, possibly mechanical, at various schools but have heard mixed reviews of them. The top schools are probably UofT and University of Waterloo, but I don't think that my average will carry me there. Which narrows me down to a few schools, McMaster, Queens, and Western. I think McMaster is alright and I've seen other forums saying that it is one of the top schools, but there are also many other forums that say their engineering program is a joke, often compared to Queens. Can anyone give me any suggestions and personal experiences regarding these universities? I know McMaster has a co-op program which I am interested in, but I know Queens also has an internship program. What are their differences? Furthermore, what are the differences in opportunities provided by each of the schools such as co-op and internship placements? I also decided to mention Western as their admission average is relatively high which should be an indicator that it is a reputable program. Thanks!

Hello all, I was wondering whether striking a match is endothermic or exothermic. This seems quite straight forward as striking a match releases heat, which is obviously exothermic. However, the process of igniting the match requires heat in itself, so I was wondering since the reaction required heat, would it be endothermic instead? I read from a post somewhere that for a reaction to be endothermic, heat is absorbed for the reaction to occur. Since heat is required for the initial ignition of the match, wouldn't that be an endothermic reaction? As well, is the bolded statement correct? Thanks

Thanks all

Why is it not possible to have the log of a negative number? Examples would be greatly appreciated. Furthermore, is it the base that cannot be negative? Like log-bx or is it that when you have a number and you log it, the number cannot be negative. log(-x)

Thanks guys, it is a quadratic. I was close!

Solve: log5(x-1) + log5(x-2) - log5(x+6) = 0 I feel like this is really simple but I cannot get the answer. This is what I attempted Since they have the same base, I multiplied the top log5[(x-1)(x-2)] - log5(x+6) = 0 This gives me log5(x2-3x+2) - log5(x+6) = 0 At this point I did not now what to do, I tried 2 methods, 1. Since they are the same base again, I divided the top this time log5(x2-3x+2) / (x+6) = 0 but I don't know how to continue 2. I moved - log5(x+6) to the other side log5(x2-3x+2) = log5(x+6) Since they have the same base I ignored them so x2-3x+2 = x+6 Which gave me x2 - 4x - 4 Which cannot be factored. Please help, thankyou

Oh I see, so they are both linear functions and I am looking for the point of intersection.

How would I actually graph it though?

Oh, log52 (x-3) = x log52 = x / (x-3) Do I use the definition of the log? 5x/(x-3) = 2 I am really confused as to what I am doing.

How would I go about graphing it? log52 (x-3) = x log52 = x (x-3) log52 = x2 - 3x what do I do after this ???? thanks.

log52 (x-3) = x One of my friends came up with this equation and asked me to graph it. I haven't seen an equation like this, and do not know how to graph using mapping notation or any other methods. Is it possible to graph this? Looking at the equation, I would assume that there can only be one answer to this because log52 can only have one answer since this means that 5x = 2 Can someone please point me in the right direction? Thanks you

Hello I'm confused as to how to do this question. 4^x + 6(4^-x) = 5 I was thinking of changing 4^x to a variable like 'a' so it would be easier to work with, but 4^-x is not the same so I can't replace that with 'a' as well. Then I though 4^-x = 1/4 ^x So 4^x + 6[(1/4)^x] = 5 But i have no idea where I'm going with this and I'm lost. Please help, thanks