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I’m not entirely sure I’ve just used them alternatively

And I didn’t think I could draw a sketch since I only have the max point, not any roots or y-intercept?
Original post by M.Johnson2111
I’m not entirely sure I’ve just used them alternatively

And I didn’t think I could draw a sketch since I only have the max point, not any roots or y-intercept?

Ok solve your quadratic in t.
image.jpgOkay would this be my final answer?
Original post by M.Johnson2111
image.jpgOkay would this be my final answer?

If you'd kept the quadratic in terms of g, then divided through by it, it would be much simpler. Dividing by g/2 would be even better.

I've not checked those numbers as you're making it too complex. But I suspect you've missed out a b^2 (at least).
(edited 3 years ago)
okay so if i leave in terms of g then divide each term by g/2

I get 2t^2 + Tt - 2T^2

Which by the formula:

t = (T +/- squareroot 15T^20 / -4
Original post by M.Johnson2111
okay so if i leave in terms of g then divide each term by g/2

I get 2t^2 + Tt - 2T^2

Which by the formula:

t = (T +/- squareroot 15T^20 / -4

Closer, what gave you that quadratic?
image.jpgimage.jpgI just want to say thank you very much for bearing with me I know I’ve made it more complex, through not understanding the question so thank you
Original post by M.Johnson2111
image.jpgimage.jpgI just want to say thank you very much for bearing with me I know I’ve made it more complex, through not understanding the question so thank you

What is the blurred stuff? Looking at your numbers you should end up with something like
0 = t^2 - 2Tt - 3T^2
Sorry i didnt realise they were blurred, okay so ive probably got a sign error,
image.jpgSo t = 5T?
Original post by M.Johnson2111
image.jpgSo t = 5T?

You really need to check/derive that quadratic. But 5T does not solve that quadratic, you can sub it in and show its not 0.
Okay im going to have another go at this tomorow and hopefully get closer, thank you for your help
Original post by M.Johnson2111
Okay im going to have another go at this tomorow and hopefully get closer, thank you for your help

Good idea. Note you should be able to directly factorize that quadratic, but starting again with a clear head is good.
HI, I ended up with total time = 3T is this similar to your result?
Of course it's 3T.

If you know it's (T,H) to the maximum, then to travel a distance 4H after the maximum is 2T, because the relationship is quadratic. So T+2T in total.
Oh wow thank you I definetly made it complicated yesterday thank you
Original post by M.Johnson2111
Oh wow thank you I definetly made it complicated yesterday thank you

When you think of the quadratic as being centred on (T,H), so complete the square, relationships like this are relatively straightforward.

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