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1. A small piece of dung is fired into the air at 30ms^-1. How long does it take to reach a height of 5m?

i know I need to use suvat equation do im using x=ut+1/2 at^2

(x)= displacment.

I presume i need to turn this into a quadratic of t in order to solve it but im not to sure how to go about this anyone give a hand.
2. What is it you don't know?
How to turn it into a quadratic, or how to solve the quadratic?
1) Sub the values you know into the formula and move x to the other side to get at² + bt + c=0
a b and c from the values given in the question and the acceleration due to gravity.
Then use the formula for solving a quadratic. [Do you know this?]
3. (Original post by Stonebridge)
What is it you don't know?
How to turn it into a quadratic, or how to solve the quadratic?
1) Sub the values you know into the formula and move x to the other side to get at² + bt + c=0
a b and c from the values given in the question and the acceleration due to gravity.
Then use the formula for solving a quadratic. [Do you know this?]
Ah sorry didn't make myself clear. Ye i just wasnt sure how to turn it into a quadratic. So to do this I use at^2=a ut=b and x =c ? If so what happens to the half?
4. Double post. The better answer is in the other one!
5. (Original post by hazbaz)
Ah sorry didn't make myself clear. Ye i just wasnt sure how to turn it into a quadratic. So to do this I use at^2=a ut=b and x =c ? If so what happens to the half?
x=ut+1/2at²

rearranges to:

1/2at²+ut-x=0

Get rid of the 1/2 by multiplying everything by 2:

at²+2ut-2x=0

(Where x is displacement, u is initial velocity, a is acceleration [deceleration] due to gravity, t is time taken to travel x distance.)

Stonebridge's point in using at²+bt+c=0 (i think) was so you could relate it to ax²+bx+c=0, where you can use x=-b +/- root (b²-4ac), all over 2a.

Sorry for the lack of latex this time.

So, a will remain acceleration, b would be 2u and c would be -2x (where x is displacement)
6. Double post. Didn't notice second topic lol.
GJ.
7. Ye thanks i got it

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