# Vertical acceleration maths Q (alevel)

I’m having difficulties with part 3 of the question.
It states that I should find the speed before P strikes the ground. However I’m unsure how I would know if the acceleration is 9.8 or -9.8?

I just chose -9.8 in my workings out because we start our journey by the particle being projected upwards, so the acceleration is -9.8ms-2.
However, will I have to change the final velocity of P when it is reaches the ground?

@TypicalNerd where have I gone wrong😭
(edited 11 months ago)
Draw a diagram with the positive direction marked on (in this case youve "chosen" upwards). s must therefore be negative as its below the initial point. a (-g) would be -9.8 and compared to u^2, v^2 must be greater as its below the inital point and youre adding on 2*9.8*0.539. Speed is positive, so take the positive root.
(edited 11 months ago)
Original post by harlz_chalamet
I’m having difficulties with part 3 of the question.
It states that I should find the speed before P strikes the ground. However I’m unsure how I would know if the acceleration is 9.8 or -9.8?

I just chose -9.8 in my workings out because we start our journey by the particle being projected upwards, so the acceleration is -9.8ms-2.
However, will I have to change the final velocity of P when it is reaches the ground?

@TypicalNerd where have I gone wrong😭

@mqb2766 has summarised it perfectly, as always.

I probably would have used a different method entirely, in which I would have considered the motion of the particle immediately after it reaches its maximum height as opposed to over the course of the full journey.

You got a correct final velocity (you should know how to work out the speed from this given the difference between a scalar and vector quantity), albeit fortuitously. I would say the only real mistakes I’ve actually noticed are that you haven’t correctly assigned signs beforehand based on direction (again, a diagram would help) and that you’ve assumed t = 0.5 s (this isn’t the length of the full journey).

Edit: I realise I perhaps haven’t given the most useful response. Notice how on your diagram that you’ve allocated the initial velocity (upwards) to be +4.9 m s^-1 and your initial acceleration (upwards to be -9.8 m s^-2? Is this not contradictory? Also, what would that suggest about the sign of the displacement?
(edited 11 months ago)
Original post by TypicalNerd
@mqb2766 has summarised it perfectly, as always.

I probably would have used a different method entirely, in which I would have considered the motion of the particle immediately after it reaches its maximum height as opposed to over the course of the full journey.

You got a correct final velocity (you should know how to work out the speed from this given the difference between a scalar and vector quantity), albeit fortuitously. I would say the only real mistakes I’ve actually noticed are that you haven’t correctly assigned signs beforehand based on direction (again, a diagram would help) and that you’ve assumed t = 0.5 s (this isn’t the length of the full journey).

Edit: I realise I perhaps haven’t given the most useful response. Notice how on your diagram that you’ve allocated the initial velocity (upwards) to be +4.9 m s^-1 and your initial acceleration (upwards to be -9.8 m s^-2? Is this not contradictory? Also, what would that suggest about the sign of the displacement?

So would If I were to make downwards my positive direction (so I have 9.8 acceleration when going down) does that mean I have to make some things, well, opposite?

Meaning, would the initial velocity instead be -4.9 because I’m taking downwards as positive

and then as @mqb2766 said, the displacement would negative because I’m going downwards instead of upwards.
and as you said, the time should be different. Should it be 1 second from the moment it’s projected to the moment it strikes the ground?
Original post by harlz_chalamet
So would If I were to make downwards my positive direction (so I have 9.8 acceleration when going down) does that mean I have to make some things, well, opposite?

Meaning, would the initial velocity instead be -4.9 because I’m taking downwards as positive

and then as @mqb2766 said, the displacement would negative because I’m going downwards instead of upwards.
and as you said, the time should be different. Should it be 1 second from the moment it’s projected to the moment it strikes the ground?

It would be more than a second, because the path of travel is not symmetrical- it falls further than it goes up, so it takes longer than 0.5 s to fall down to the ground.

But yes, u = -4.9.