Projectiles M1

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Original post by user342

I think you need to check your +ve and -ve signs, ie if u is upwards, then u must be negative, and if downwards is positive, then acceleration is +9.8. I get u=14m/s though not 0.14 which is weird.

Agree with that (14 m/s).
If u was 0.14, thats almost zero and the time to fall ~10m under gravity would be ~1.4s, not ~3.5s

(edited 2 years ago)

Reply 3

OP

alright i understood this part. i get 14 ms^1 too.
i believe, that i still don't quite have a clear understanding in this.
so i have some questions,

*the time 25/7 s, is the entire time for the ball to fall from A to when it hits the ground, right?

*about the acceleration, the ball is thrown vertically upwards, and so i took the acceleration as negative.
i don't understand why the accel. should be +9.8 ms^-2.

Reply 4

Original post by Aleksander Krol

alright i understood this part. i get 14 ms^1 too.
i believe, that i still don't quite have a clear understanding in this.
so i have some questions,

*the time 25/7 s, is the entire time for the ball to fall from A to when it hits the ground, right?

*about the acceleration, the ball is thrown vertically upwards, and so i took the acceleration as negative.
i don't understand why the accel. should be +9.8 ms^-2.

If positive is upwards (it does not matter what the initial direction of motion is),
s = -12.5
a = -9.8
u = 14
...

If downwards is positive, flip the sign on all these terms.

(edited 2 years ago)

Reply 5

OP

Original post by mqb2766

If positive is upwards (it does not matter what the initial direction of motion is),
s = -12.5
a = -9.8
u = 14
...

If downwards is positive, flip the sign on all these terms.

and about the time? is 25/7 s the time it took to fall from A to the ground?

Reply 6

a

Original post by Aleksander Krol

and about the time? is 25/7 s the time it took to fall from A to the ground?

Thats what the question said and I'm sure you could easily verify by sticking u=14 (upwards positive) into s=ut+1/2at^2 if you were unsure.

Reply 7

OP

Original post by mqb2766

a

Thats what the question said and I'm sure you could easily verify by sticking u=14 (upwards positive) into s=ut+1/2at^2 if you were unsure.

okay, so from what i understood is, that the vectors cancel out, and the only vector left is the yellow arrow pointing downwards (+) sign, cause i took downwards to be positive.
so since the the yellow arrow is pointing down, the acceleration is also positive.

in this case, the displacement will be 0, cause the vectors cancel out and the acceleration will be negative, cause it's thrown vertically upwards and i took upwards to be negative.

here, the displacement will be negative 2m and there' just only one vector, so nothing to cancel out.
and the acceleration will be negative, cause it's thrown vertically upwards and i took the upwards to be negative.

(edited 2 years ago)

Reply 8

Im presuming your +/- arrows indicate direction? If so, they don't cancel, they're used to indicate the direction of positive and negative quantities. In your original diagram (positive up), both s (ground has negative dispacement compared to origin) and g (accelerating downwards) will be negative and u is positive as its thrown upwards.

(edited 2 years ago)

Reply 9

OP

Original post by mqb2766

Im presuming your +/- arrows indicate direction? If so, they don't cancel, they're used to indicate the direction of positive and negative quantities. In your original diagram (positive up), both s (ground has negative dispacement compared to origin) and g (accelerating downwards) will be negative and u is positive as its thrown upwards.

those arrows are indicating displacement signs. i mean, the displacement cancels and that's how we're left with 12.5 m.
that's the reason why we aren't considering the greatest height it reached from A + the greatest height to the point A + A to the ground (which is 12.5m from A to the ground) right?

Reply 10