I don't get what you mean sorry.

y=xtan30 - 3/40x^2

Sub 0.6 in and multiply by 40.

24 = 40xtan30 - 3x^2
Make x^2 positive
3x^2-40xtan30 +24 =0
Solve quadratic.

Reply 7

Zacken

Original post by Super199

24 = 40xtan30 - 3x^2
Make x^2 positive
3x^2-40xtan30 +24 =0
Solve quadratic.

Urgh, I'm being an idiot. Your mistake was in the first part.

You're dividing by (for the x^2 term)

\cos^2 30^{\circ} = \frac{3}{4}

so you should multiply by

\frac{4}{3}

. You're multiplying by

\frac{3}{4}

.

That is:

\frac{4.9x^2}{49 \cos^2 30^{\circ}} = \frac{2x^2}{15}

Reply 8

Super199

Original post by Zacken

Urgh, I'm being an idiot. Your mistake was in the first part.

You're dividing by (for the x^2 term)

\cos^2 30^{\circ} = \frac{3}{4}

so you should multiply by

\frac{4}{3}

. You're multiplying by

\frac{3}{4}

.

That is:

\frac{4.9x^2}{49 \cos^2 30^{\circ}} = \frac{2x^2}{15}

No worries :smile:

Do you mind helping with the third part. What does it mean by 'and the particle is rising' what does that effect?

Reply 9

Zacken

Original post by Super199

No worries :smile:

Do you mind helping with the third part. What does it mean by 'and the particle is rising' what does that effect?

It just means that there are two moments where y=0.6, one when the particle is going up, passes the y=0.6 mark and continues going up to the max height before falling back down, passing the y=0.6 mark again and then falling down to the ground. The question is specifying to work with the first moment when the particle crosses the y=0.6 mark on its way up.

Reply 10

Super199

Original post by Zacken

So it has nothing to do with ii?

Reply 11

Zacken

Original post by Super199

So it has nothing to do with ii?

Yep. It's just normal SUVAT.

(Although there is an alternative method where you can differentiate the answer to (i), blah blah but it's more complicated)

Reply 12

Super199

Original post by Zacken

Yep. It's just normal SUVAT.

(Although there is an alternative method where you can differentiate the answer to (i), blah blah but it's more complicated)

yh fair enough. So what was the point of putting that particle rising bs. Like that ****ed with me and I thought I had to use x= 1.73 from the previous part or some ****.

Reply 13

Zacken

Original post by Super199

yh fair enough. So what was the point of putting that particle rising bs. Like that ****ed with me and I thought I had to use x= 1.73 from the previous part or some ****.

Like I said, it specifies that it wants the direction when the particle is going up and not at the second moment when the particle is going down. Because there are two points on the parabolic trajectory where y=0.6.

Reply 14

Super199

Original post by Zacken

would you use u=7sin30 if the particle was going down? Or is that only for up

Reply 15

Zacken

Original post by Super199

would you use u=7sin30 if the particle was going down? Or is that only for up

You'd use it for both.

Reply 16

Super199

Original post by Zacken

You'd use it for both.

Cheers.
With q1 I never understand how these energy qs work.
Why don't we consider the resistance to motion * 30 as an energy?

Reply 17

Zacken

Original post by Super199

Cheers.
With q1 I never understand how these energy qs work.
Why don't we consider the resistance to motion * 30 as an energy?

Where did 30 come from?

Reply 18

Super199

Original post by Zacken

Where did 30 come from?

Sorry 20m. As that is the work done by the resistance?

Reply 19

Zacken

Original post by Super199

Sorry 20m. As that is the work done by the resistance?

Yes, it's the work done against resistance, but the energy of the sledge and load has nothing to do with the resistance of the motion, the energy of the sledge and load is purely K.E - G.P.E. The work done against resistance is used in the second part.