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M2 particles urgent help

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?

Reply 1

ghostwalker

Original post by dskinner

What is this meant to be? In red.

Reply 2

dskinner

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

Reply 3

ghostwalker

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.

Reply 4

dskinner

Original post by ghostwalker

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

Reply 5

ghostwalker

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.

And add.

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.

Reply 6

dskinner

Original post by ghostwalker

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

Reply 7

ghostwalker

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

Which is about right.

Reply 8

dskinner

Original post by ghostwalker

Which is about right.

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?

Reply 9

ghostwalker

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

Reply 10

dskinner

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?

Reply 11

ghostwalker

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.

Reply 12

dskinner

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.