jamestg Maths help

OP

First question - what on earth do I do for part (c)?

Also is:
part (a) - k^2+2k+9
part (b) - (k+1)^2 +8

Reply 2

math42

20

Original post by jamestg

First question - what on earth do I do for part (c)?

Also is:
part (a) - k^2+2k+9
part (b) - (k+1)^2 +8

Pretty sure those are right. Remember the rules for a discriminant. If it is less than zero, you have no real roots; if equal to zero you have a single root (or two "repeated real roots"); if more than zero you have two distinct real roots.

(edited 8 years ago)

Reply 3

OP

Original post by 1 8 13 20 42

Pretty sure those are right. Remember the rules for a discriminant. If it is less than zero, you have no real roots; if equal to zero you have a single root (or two "repeated real roots" :wink:

; if more than zero you have two distinct real roots.

Thank you!

So to answer the question, I just need to substitute a value in for k?

Reply 4

math42

20

Original post by jamestg

Thank you!

So to answer the question, I just need to substitute a value in for k?

Nah because that would only prove it for one value of k. You need to show that, no matter what k is, (k + 1)^2 + 8 will be greater than zero. Consider the (k + 1)^2 part first.

Reply 5

OP

I've done pretty well to not have anything to ask for almost two months!

I'm stuck on an M1 question at the moment...

An object is projected vertically upwards from ground level with a speed of 30m/s.
a) find the maximum height reached by the object
b) find the time it takes the object to land on the ground

For part a I got 45.9m and part b I got 5.4secs. Is this correct? I've honestly got no idea...

Reply 6

Original post by jamestg

I've done pretty well to not have anything to ask for almost two months!

I'm stuck on an M1 question at the moment...

An object is projected vertically upwards from ground level with a speed of 30m/s.
a) find the maximum height reached by the object
b) find the time it takes the object to land on the ground

For part a I got 45.9m and part b I got 5.4secs. Is this correct? I've honestly got no idea...

Part a is fine.

Reply 7

Original post by jamestg

First question - what on earth do I do for part (c)?

Also is:
part (a) - k^2+2k+9
part (b) - (k+1)^2 +8

Find the discriminant. b^2 - 4ac. a = 1, b = k +1 (or whatever it is), and c = 1. Then see if it has any real roots.

Reply 8

Original post by jamestg

I've done pretty well to not have anything to ask for almost two months!

I'm stuck on an M1 question at the moment...

An object is projected vertically upwards from ground level with a speed of 30m/s.
a) find the maximum height reached by the object
b) find the time it takes the object to land on the ground

For part a I got 45.9m and part b I got 5.4secs. Is this correct? I've honestly got no idea...

Don't think the second part looks right, working?

Reply 9

Original post by ravioliyears

Find the discriminant. b^2 - 4ac. a = 1, b = k +1 (or whatever it is), and c = 1. Then see if it has any real roots.

That was two months ago and he already got an answer. :tongue:

Reply 10

Original post by jamestg

I've done pretty well to not have anything to ask for almost two months!

I'm stuck on an M1 question at the moment...

An object is projected vertically upwards from ground level with a speed of 30m/s.
a) find the maximum height reached by the object
b) find the time it takes the object to land on the ground

For part a I got 45.9m and part b I got 5.4secs. Is this correct? I've honestly got no idea...

I'm not sure, but I think you would do 30sin90 to find Uy, and then to work out t you do 2Uy/a (a being 9.81 m/s)???

Reply 11

Original post by Zacken

That was two months ago and he already got an answer. :tongue:

Oh lord was it? Lol it was trending so I just answered it :tongue:

Reply 12

Original post by ravioliyears

I'm not sure, but I think you would do 30sin90 to find Uy, and then to work out t you do 2Uy/a (a being 9.81 m/s)???

30 sin 90 = 30 because the velocity is already completely vertical. :-)

Reply 13

OP

Original post by Zacken

Don't think the second part looks right, working?

Reply 14

OP

Original post by ravioliyears

I'm not sure, but I think you would do 30sin90 to find Uy, and then to work out t you do 2Uy/a (a being 9.81 m/s)???

Original post by Zacken

30 sin 90 = 30 because the velocity is already completely vertical. :-)

Why is sin being used?! This hasn't come up in any of our lessons!

Reply 15

Original post by jamestg

I'm not too sure what you're doing, but what do you get if you set s=0, u=30, a=-9.8, t=?

You should get a quadratic in t with a solution t=0 and the other solution your answer.

In your working, you seem to have put s=+ve value, but s = -ve because you're going from top to down and that's negative.

Reply 16

Original post by jamestg

Why is sin being used?! This hasn't come up in any of our lessons!

I'm so sorry to confuse you - I was trying to work it out using projectiles. Just ignore that all

(edited 8 years ago)

Reply 17

Original post by jamestg

Why is sin being used?! This hasn't come up in any of our lessons!

Ignore it.

Reply 18