kawaiipotato
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- Apr 28, 2015
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- HSC
- 2015
Anyone got q21? The height of the wall? I found grqdient of line and subbed into y = ut +1/2 a_y t^2 where u = 0 and t = gradient. Lol dunno if its correct
Pretty sure it's incorrect, but I did it anyway.Anyone got q21? The height of the wall? I found grqdient of line and subbed into y = ut +1/2 a_y t^2 where u = 0 and t = gradient. Lol dunno if its correct
I got 2.5 metresAnyone got q21? The height of the wall? I found grqdient of line and subbed into y = ut +1/2 a_y t^2 where u = 0 and t = gradient. Lol dunno if its correct
It's correct, gradient is sx/id which gives time, then you can sub it into sy = 1/2at^2
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I felt like it was pretty easy except for 1 or 2 questions but I made a lot of stupid errors I think because I just found that it look really long so I didn't have time to check.
Sweet. I was put off by a lot of the marks given - I feel like it was a very generous exam.
Ah cool. What was the numerical value? I think i remember it being 4. Something?It's correct, gradient is sx/ux which gives time, then you can sub it into sy = 1/2at^2
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Well I got 2.5 but not sure if it's right.Ah cool. What was the numerical value? I think i remember it being 4. Something?
My gradient was like 0.86Well I got 2.5 but not sure if it's right.
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Mine was whatever gives 2.5My gradient was like 0.86
Yeah I had it as that, not too sure whether it's correct. Also had an outlier that I didn't include.did your line of best fit start from 0?
i did it from first value to last and not including an outlierdid your line of best fit start from 0?
mine excluded the outlier but it didnt touch the first or last value lol, it was in the middle of them and started at 0,0i did it from first value to last and not including an outlier
brb witcher 3
Using the given data or the line of best fit?Didn't you just have to use the Vertical displacement formula? I got 2.5m as well
I used the data. More specifically, the horizontal displacement and the velocity to find the time and then subbed that into the vertical displacement formula. Maybe I switched off at that stage but yeah that's what I didUsing the given data or the line of best fit?
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That's right but I think they wanted us to use the gradient of our line of best fit because it's more accurate or something but not sure. Either way I got 2.5 because my line passed through the final point anyway.I used the data. More specifically, the horizontal displacement and the velocity to find the time and then subbed that into the vertical displacement formula. Maybe I switched off at that stage but yeah that's what I did