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Kinematics SUVAT question help AS????

Hi, i need help on the following:

A particle is projected vertically upwards with a speed of 10ms-1 from a point 4m above the ground.

i - Find the maximum height reached above the ground of the particle

I did this correctly and found it to be 9.1m.

ii- Find the speed of the particle when it hits the ground - this is the bit I'm confused on. I found this question in a video but for the values of suvat he says that u=10? surely u=0 because when it reached its peak and begins to fall its initial velocity is 0. so s=-4, u=0, v=?, a=9.8, t=? Furthermore, the guy set upwards to be negative for the first question and so a=-9.8, but the the second part he wrote a=-9.8? I'm quite confused on all of this.

many thanks
(edited 6 years ago)

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Anyone
Original post by Bertybassett
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But when the particle as at its peak, then it has to cover a displacement of -9.1m with initial speed 0, does it not? Why are you saying the s=-4?

I do not know what approach 'the guy' is using from the way you describe it TBH, but you can alternatively say that you want the time when your particle is displaced by -4 vertically, so that it is on the ground. Setting s=−4s=-4, u=10u=10, a=−9.8a=-9.8 and find the positive value of time from s=ut+12at2s=ut+\frac{1}{2}at^2 will give the amount of time since launch that the particle takes to come back to ground. Then using v=u+atv=u+at, and taking the magnitude of it will give you the required speed. This is what I'd image 'this guy' was doing.
Original post by RDKGames
But when the particle as at its peak, then it has to cover a displacement of -9.1m with initial speed 0, does it not? Why are you saying the s=-4?

I do not know what approach 'the guy' is using from the way you describe it TBH, but you can alternatively say that you want the time when your particle is displaced by -4 vertically, so that it is on the ground. Setting s=−4s=-4, u=10u=10, a=−9.8a=-9.8 and find the positive value of time from s=ut+12at2s=ut+\frac{1}{2}at^2 will give the amount of time since launch that the particle takes to come back to ground. Then using v=u+atv=u+at, and taking the magnitude of it will give you the required speed. This is what I'd image 'this guy' was doing.


Hi thanks for the reply, the displacement is -4 i think because the question says that the particle's starting point if 4m above the ground, but it falls to the ground 4m below its starting point? could you please explain how you got the value of u to be 10, as surely when the particle reaches it's peak it's veloicty is 0 before acceleterating downwards? also, why did you set a to -9.8, when the particle is coming back towards earth and the man said that coming down was positive g and upwards was negative? thanks
(edited 6 years ago)
Original post by Bertybassett
Hi, i need help on the following:

A particle is projected vertically upwards with a speed of 10ms-1 from a point 4m above the ground.

i - Find the maximum height reached above the ground of the particle

I did this correctly and found it to be 9.1m.

ii- Find the speed of the particle when it hits the ground - this is the bit I'm confused on. I found this question in a video but for the values of suvat he says that u=10? surely u=0 because when it reached its peak and begins to fall its initial velocity is 0. so s=-4, u=0, v=?, a=9.8, t=? Furthermore, the guy set upwards to be negative for the first question and so a=-9.8, but the the second part he wrote a=-9.8? I'm quite confused on all of this.

many thanks



For maximum height, we take v to equal 0, and a to equal -9.81 (as you're going against the force of gravity), and u to be 10, plonk that into SUVAT and see what you obtain.

Find the speed it goes when it falls back down, well a nifty way to do it is figure out your time in the equation RDK kindly mentioned previously, and then simply plugging that into SUVAT you obtain a value of V.
Original post by AryanGh
For maximum height, we take v to equal 0, and a to equal -9.81 (as you're going against the force of gravity), and u to be 10, plonk that into SUVAT and see what you obtain.

Find the speed it goes when it falls back down, well a nifty way to do it is figure out your time in the equation RDK kindly mentioned previously, and then simply plugging that into SUVAT you obtain a value of V.


thanks, could you explain why 9.8 is negative though? if the object is falling wont it be positive?
Original post by Bertybassett
thanks, could you explain why 9.8 is negative though? if the object is falling wont it be positive?


As we're taking the maximum height here, so the objects maximum displacement from the ground, we split the motion in half. (It is assumed that with negligible air resistance, the time to go up will equal the time to go down).

You should know that projectile motion can be modelled as a parabola, or a curve.

At maximum height, it has reached it's highest height, it cannot go up further, from that point onwards, it then begins to fall.

So at the first half, we're throwing an object upwards to find maximum height, against the force of gravity, 9.81 metres per second, per second, which acts downwards on a body, thus we take gravity in this instance, and thus a, to be -9.81 as it's going up at first and not down.
Original post by Bertybassett
Hi thanks for the reply, the displacement is -4 i think because the question says that the particle's starting point if 4m above the ground, but it falls to the ground 4m below its starting point?


That's right, however YOU are considering motion starting from the max height, therefore the displacement needs to be -9.1 with initial velocity 0 if you want to do it your way.

could you please explain how you got the value of u to be 10, as surely when the particle reaches it's peak it's veloicty is 0 before acceleterating downwards?


Because I am starting from it's initial point which is 4m above ground, NOT from its peak height which is 9.1m! At the initial point, u=10.

also, why did you set a to -9.8, when the particle is coming back towards earth and the man said that coming down was positive g and upwards was negative? thanks


Because we are saying whatever is going in the direction UP is positive, therefore, as acceleration is acting down, we say it is negative.
I don't know why 'the man' said otherwise, but it's best if we are consistent throughout our working here. Acceleration due to gravity does not magically change its direction whenever we want it to. We need to agree that UP means +ve and DOWN means -ve and stick with it, perhaps 'the man' swapped these around - which quite frankly, I don't prefer to do unless I have to, and here it's not necessary.

------------------------------------------------

All in all, there are two ways to consider part (ii)

YOUR WAY: Consider motion of the particle starting from its peak height, with u=0u=0 and height 9.1m above ground, moving downwards at a=−9.8a=-9.8. To reach the ground, it must travel 9.1m, therefore we say s=−9.1s=-9.1 (and it's -ve because we want it displace DOWNWARDS from its max height). This yields equation −9.1=−12(9.8)t12-9.1=-\frac{1}{2}(9.8)t_1^2 from which the final velocity is given by v=−9.8t1v=-9.8t_1 which is negative, and that's fine, because the particle is moving DOWNWARDS. The speed is given by the magnitude of this velocity.

MY WAY: Consider motion of the particle starting from its initial position with u=10u=10 and height 4m above ground, moving downwards at a=−9.8a=-9.8. To reach the ground, it must travel 4m, therefore we say s=−4s=-4. This yields −4=10t2−12(9.8)t22-4=10t_2-\frac{1}{2}(9.8)t_2^2 and so we use again v=10−9.8t2v=10-9.8t_2 which gives the exact same answer as in YOUR WAY. Taking the magnitude yields the same speed.


Get it?
(edited 6 years ago)
A more simple way of looking at it is by setting the variables respectively:
S - 9.1 + 4
U - 0
V - What you're trying to find
A - 9.8
T - unnecessary information.

You have enough information after completing the first part to do it this way, and it is much easier to think of it like this rather than using negative displacements. Just my opinion though :smile:
Reply 9
Avatar for Pastelx
Pastelx
18
Original post by Bertybassett
thanks, could you explain why 9.8 is negative though? if the object is falling wont it be positive?


Just bare in mind there is no set direction to take as positive when it comes to suvat, as long as you are consistent

e.g. my physics teacher takes downwards as positive since that generally means less negatives to work with whereas I usually take upwards as positive since I visualise with graphs how the y-axis is positive when going upwards.

As long as displacement, initial/final velocity and acceleration take the same/opposite signs where appropriate, you’ll get the answer either way (due to cancelling out of signs)
Original post by _RobbieL_

S - 9.1 + 4


At what point is the displacement 13.1?
Reply 11
Avatar for thiccc
thiccc
17
surely you can use v^2=u^2 * 2as again.

V=?
U=0
a=9.81
s=9.1

therefore v = 13.3m/s.

Another way to do it (non suvat) would be PE max = KE max.

mgh=1/2mv^2.

The m cancels and you are pretty much left with the same suvat equation.
Original post by RDKGames
At what point is the displacement 13.1?


S can be either displacement, or distance. If you think of it in terms of distance, the ground to the peak of the projectile is 13.1m.
Original post by _RobbieL_
S can be either displacement, or distance. If you think of it in terms of distance, the ground to the peak of the projectile is 13.1m.


ss is defined to be displacement from an initial position because distance traveled is not a vector quantity therefore it does not work with the rest of the vector quantities like U (initial velocity) and V (final velocity) as well as A (acceleration)

Clearly, the dimensions of s=12(u+v)ts=\frac{1}{2}(u+v)t would be inconsistent if ss was taken to be a scalar quantity, ie distance traveled.
Original post by RDKGames
ss is defined to be displacement from an initial position because distance traveled is not a vector quantity therefore it does not work with the rest of the vector quantities like U (initial velocity) and V (final velocity) as well as A (acceleration)

Clearly, the dimensions of s=12(u+v)ts=\frac{1}{2}(u+v)t would be inconsistent if ss was taken to be a scalar quantity, ie distance traveled.


I promise you if you did the calculation using my variables the answer you get for V would be correct.
Original post by _RobbieL_
I promise you if you did the calculation using my variables the answer you get for V would be correct.


v2=u2+2asv^2=u^2+2as yields v≈16v \approx 16 which is incorrect with s=13.1s=13.1
Original post by RDKGames
v2=u22asv^2=u^2 2as yields v≈16v \approx 16 which is incorrect.


It gets 13.3m/s when s is 9.1, so if it was 13.1 I think approx 16 sounds about right.
(edited 6 years ago)
Original post by _RobbieL_
It gets 13.3m/s as thicc said above.


Because they used s=9.1s=9.1 which is the correct displacement to use.
Original post by RDKGames
Because they used s=9.1s=9.1 which is the correct displacement to use.


Mate you’re literally forgetting that the particle was projected from 4m above the ground, therefore there is an extra 4 metres to travel for it to reach the ground from it’s peak. I’m not lying.
Reply 19
Avatar for thiccc
thiccc
17
Original post by _RobbieL_
Mate you’re literally forgetting that the particle was projected from 4m above the ground, therefore there is an extra 4 metres to travel for it to reach the ground from it’s peak. I’m not lying.


s=9.1 already includes the 4m in the calculation.

quick maths

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