4:28 AM
@Pizza Thanks :-)
@BinkyMcSquigglebottom That's solving for š” using the height change then using x = vt

5 hours later…
9:09 AM
@JohnRennie I have seen the solutions to this problem but I wanted to know if we can solve this question using a phasor diagram? As in, can we draw phasor diagram for when the 2 forces acting on a partile aren't linear?
Hi :-)
I don't think this would work. The trouble is that we don't have a linear system here i.e. you can't just add the two displacements.
I would do it by splitting s(t) into x and y components, and then adding the x component to the first motion, which I'm guessing is the solution you have seen already.
9:27 AM
@JohnRennie Thanks :-)
10:09 AM
@JohnRennie for Binky's exercise, that's what you mean, right?
$v_{0x}=v_0\cdot\cos(\theta)$
$v_{0y}=v_0\cdot\sin(\theta)$
$d_x =v_{0x}\cdot t \Rightarrow d_x =v_0\cdot\cos(\theta)\cdot t \Rightarrow t=\frac{d_x}{v_0\cdot\cos(\theta)}$ $\quad(1)$
$d_y = v_{0y}\cdot t-\frac{1}{2}gt^2\Rightarrow d_y=v_0\cdot\sin(\theta) \cdot \frac{d}{v_0\cdot\cos(\theta)} - \frac{1}{2}g\cdot\left(\frac{d}{v_0\cdot\cos(\theta)}\right)^2$ $\quad(2)$$t=\frac{9400m}{v_0\cdot\cos(35°)}\tag{1}$$$$-3300m=v_0\cdot\sin(35°)\cdot\frac{9400m}{v_0\cdot\cos(35°)}- \frac{1}{2}g\left(\frac{9400m}{v_0\cdot\cos(35°)}\right)^2\tag{2}$$ @Pizza Yes There are several approaches to the problem, but they all basically involve x(t) = 9400 and y(t) = -3300 at the same time š”. 10:41 AM $$-3300m = 9400m \cdot \tan(35°) - \frac{1}{2} 9.81m/s^2\cdot\frac{(9400m)^2}{v_0^2 \cos^2(35°)}$$$$9881m = 4,905m/s^2\cdot\frac{88360000m^2}{v^2_0 \cdot 0,671}$$$$v^2_0\cdot9881m=4,905m/s^2\cdot \frac{88360000m^2}{0,671}$$$$v^2_0=4,905m/s^2\cdot\frac{88360000m^2}{9881m\cdot0,671}=\frac{433405800m^3/s^2}{6630,151m}\approx65368,9m^2/s^2$$$$v_0=\sqrt{65368,9m^2/s^2}\approx 255m/s$$ Yes, that's the answer I got. if we wanted to find the flight time, we can do it like this, right? $$t_f=\frac{d_x}{v_0\cos(\theta)}=\frac{9400m}{255m/s\cdot\cos(35°)}=45s$$ where$t_f=$flight time and$d_x =$the distance on the$x$axis @JohnRennie however, without a calculator the calculation of the first point was a little complicated š @JohnRennie but why if I try to find$v_0$from that formula do I find it different? @Pizza Yes, I did it using Google Sheets. When I have involved calculations I often lay them out in a spreadsheet to keep then clear. @BinkyMcSquigglebottom Hi :-) @JohnRennie nice idea! hi @BinkyMcSquigglebottom 10:56 AM 23 hours ago, by Binky McSquigglebottom 9400=x^2Sin[2*35°]/2*9,81(1+Sqrt[1+2*9,81*3300/x^2(sin(2*35°))^2)] There is a possible error in the equation. I don't know exactly how Wolfram would parse it, but 2*9.81 should be in brackets. 9400=x^2Sin[2*35°]/(2*9,81) * (1+Sqrt[1+2*9,81*3300/x^2(sin(2*35°))^2)]) 18 hours ago, by Pizza $$x_G = \frac{v^2_0\sin 2\theta}{2g} \left( 1 + \sqrt{1 + \frac{2g \ y_0}{v^2_0 \sin^2 \theta} \ }\right)$$ This is the equation the same is not found š­ Also note that yā needs to be negative because we are taking the upwards direction to be positive so a height below the starting point is negative. @BinkyMcSquigglebottom I'm not sure what that means ... nothing changes I still can't find the right solution Do you want to go through how I did the calculation? I'm only here for another 20 minutes. After that we'll have to continue tomorrow. 11:13 AM $$9400 \text{ m} = \frac{v^2_0 \cdot 0.9397}{2 \cdot 9.8 \text{ m/s}^2} \left( 1 + \sqrt{1 + \frac{2 \cdot 9.8 \text{ m/s}^2 \cdot (-3300 \text{ m})}{v^2_0 \cdot \sin^2(35°)}} \right)$$ @JohnRennie it should look like this, using @BinkyMcSquigglebottom formula, right? I guess so. I used a different method and I have not checked their equation. But you got the correct answer so it must be right :-) @JohnRennie I think he wants to use this equation but he said "I still can't find the right solution" he wasn't referring to your method (the one I used previously) Ah, OK, you used the same method as me then? yes the formula that I sent now is the @BinkyMcSquigglebottom one I'd have to go through their equation to check it but I'm not at my desk so I'll have to do it later. 11:19 AM In my opinion this is much more complicated, it is better to use your method on wikipedia it says that the @BinkyMcSquigglebottom formula is: the case in which the starting height is not zero. All we need to do is reuse the function y(x) by adding the constant$y_0 \ne 0$. I agree! :-) $$9400 \text{ m} = \frac{v^2_0 \cdot 0.9397}{19.6 \text{ m/s}^2} \left( 1 + \sqrt{1 - \frac{64680m^2/s^2}{v^2_0 \cdot 0,329}} \right)$$ there are 2$v_0$variables, I don't know how to get out of it I guess that's why they used Wolfram. ye... It's not that hard but it's messy. 11:27 AM I'd say work smarter, not harder at this point Rearrange and square both sides and it turns into a quadratic in vā² actually for$v_0\$ there are no solution
I guess the equation must be wrong ...
Anyhow, I need to now. Bye :-)
@JohnRennie If you do it again from the beginning, let me know
@JohnRennie bye:)