Thread: SUVAT - Straight forward question... quadratics

Hi Guys,

Here is a pretty straight forward question.

The only problem is, I got confused as to what happens with - and + numbers when they are squared etc...

Could anybody kindly guide me through the following problem and explain step by step what is taking place?

Ok here goes:

Q. A body is projected vertically upwards from ground level at a speed of 24.5ms-¹. Calculate the length of time that the stone is at least 19.6m above ground level.

I understand that if I use SUVAT, s = 19.6, u = 24.5, a = -9.8ms²

Therefore I must use: s = ut + 1/2 at²

Therefore: 19.6 = 24.5(t) + 1/2 -9.8(t&#178

Therefore: 19.6 = 24.5(t) + -4.9(t&#178

Therefore: ??? Now i'm unsure how to make t the subject, what happens with the +-4.9 ???

Originally Posted by c00ky

I understand that if I use SUVAT, s = 19.6, u = 24.5, a = -9.8ms²

Therefore I must use: s = ut + 1/2 at²

Therefore: 19.6 = 24.5(t) + 1/2 -9.8(t&#178

Therefore: 19.6 = 24.5(t) + -4.9(t&#178

Therefore: ??? Now i'm unsure how to make t the subject, what happens with the +-4.9 ???

adding a negative means subtracting it. But that doesn't matter because what you do is subtract 19.6 from both sides to get:

Use the quadratic formula:

Substitute:

Solving that will give you the two times when the ball is 19.6m above ground. Then you just have to find the difference between them

Would it not be easier / possible to transpose the formula to make t the subject?

Originally Posted by c00ky

Would it not be easier / possible to transpose the formula to make t the subject?

The quadratic formula makes the subject, if there is a better way then I don't see it for your particular problem.

Originally Posted by c00ky

Would it not be easier / possible to transpose the formula to make t the subject?

Originally Posted by Quick

The quadratic formula makes the subject, if there is a better way then I don't see it for your particular problem.

The quadratic formula is the only method to use here, unless you want to "complete the square" (which is how the quadratic formula is derived in the first place.)

-Dan