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# M2 particles urgent help watch

1. A particle P is projected from a point A with speed 32m/s at an angle of elevation alpha where sin alpha=3/5 The point O is on horizontal ground with O vertically below A and OA=20m The particle P moves freely under gravity and passes a point B which is 16m above ground before reaching the ground at point C

Calculate the time of flight from A to C

the distance OC

the speed of P at B

the angle that the velocity of P at B makes with the horizontal

resolving vertically i got u=19.2 v=0 a=-9.8 and t=t
i found t to be 1.96 to 2 dp

the i did u=-19.2 s=20 t=t and a=-9.8

where did i go wrong?
2. (Original post by dskinner)
A particle P is projected from a point A with speed 32m/s at an angle of elevation alpha where sin alpha=3/5 The point O is on horizontal ground with O vertically below A and OA=20m The particle P moves freely under gravity and passes a point B which is 16m above ground before reaching the ground at point C

Calculate the time of flight from A to C

the distance OC

the speed of P at B

the angle that the velocity of P at B makes with the horizontal

resolving vertically i got u=19.2 v=0 a=-9.8 and t=t
i found t to be 1.96 to 2 dp
What is this meant to be? In red.
3. (Original post by ghostwalker)
What is this meant to be? In red.
lemme change that to u=-19.2 s=-20 t=t a=-9.8

where i'm doing suvat on the bit there the particle falls down and i'm using u=-19.2 because there are 2 places where u=19.2
4. (Original post by dskinner)
lemme change that to u=-19.2 s=-20 t=t a=-9.8

where i'm doing suvat on the bit there the particle falls down and i'm using u=-19.2 because there are 2 places where u=19.2
I assume this is the first part for the time of flight from A to C.

I don't understand your reason for using u=-19.2.

If we assign upwards as positive, then u=19.2, and you have s and a, correct, being negative.
5. (Original post by ghostwalker)
I assume this is the first part for the time of flight from A to C.

I don't understand your reason for using u=-19.2.

If we assign upwards as positive, then u=19.2, and you have s and a, correct, being negative.
Yes so calculate the time from u=19.2 to v=0 where it stops momentarily i used u=19.2 v=0 a=-9.8 and t=t

then when the particle comes back down again i used u=-19.2 s=-20 t=t a=-9.8
6. (Original post by dskinner)
Yes so calculate the time from u=19.2 to v=0 where it stops momentarily i used u=19.2 v=0 a=-9.8 and t=t

then when the particle comes back down again i used u=-19.2 s=-20 t=t a=-9.8
OK, I think I can follow what you're doing now.

You've calculated the time from the start to the max height. Which you then need to double to get the time until it's back to the same height it started from.

Then you're calculating the time from that start height to when it reaches the ground, in which case your u=-19.2 s=-20 t=t a=-9.8 would be correct.

Seems a rather roundabout way to do it. I would do it in one, motion from A to C.
s=-20 u=19.2 a=-9.8 t=t. Using s=ut+1/2 at^2, to get a quadratic in t, and solve with formula.
7. (Original post by ghostwalker)
OK, I think I can follow what you're doing now.

You've calculated the time from the start to the max height. Which you then need to double to get the time until it's back to the same height it started from.

Then you're calculating the time from that start height to when it reaches the ground, in which case your u=-19.2 s=-20 t=t a=-9.8 would be correct.

Seems a rather roundabout way to do it. I would do it in one, motion from A to C.
s=-20 u=19.2 a=-9.8 t=t. Using s=ut+1/2 at^2, to get a quadratic in t, and solve with formula.
didn't know you could do that....

but in the end i got 1.96x2 + 0.86 which comes to a grand total of 4.8s
8. (Original post by dskinner)
didn't know you could do that....

but in the end i got 1.96x2 + 0.86 which comes to a grand total of 4.8s
9. (Original post by ghostwalker)
so the second part for the distance from O to C

i resolved horizontally

u=25.6 a=0 t=4.8 s=s

but i get a huge number is that right?
10. (Original post by dskinner)
so the second part for the distance from O to C

i resolved horizontally

u=25.6 a=0 t=4.8 s=s

but i get a huge number is that right?

"Huge" is rather vague. s= 25.6 x 4.8
11. (Original post by ghostwalker)
"Huge" is rather vague. s= 25.6 x 4.8
ok so 122m

the for speed of P at B

i got u=-19.2 s=-4 v=v and a=-9.8 is that ok?
12. (Original post by dskinner)
ok so 122m

the for speed of P at B

i got u=-19.2 s=-4 v=v and a=-9.8 is that ok?
That's the vertical velocity, but you need to combine it with the horizontal to get the speed - pythagoras.

This is something that should be covered in your textbook.
13. (Original post by ghostwalker)
That's the vertical velocity, but you need to combine it with the horizontal to get the speed - pythagoras.

This is something that should be covered in your textbook.
ah... well i don't have a textbook just yet... book return day for the U6th is soon
thanks
14. Just set it out like this and these problems should be pretty straightforward.

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