physics question, i am quite stupidWatch

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Thread starter 9 months ago
#1
A ball is dropped from a height of 3.0 m. How long does it take to reach the ground?

I am currently practising for GCSE mock exams, I have been procrastinating a lot and already struggling on the first multiple choice question. Please help, thanks a lot BTW, there might be something wrong with the question, as they did not tell any data, just the height (distance)...
Last edited by giolaw; 9 months ago
0
9 months ago
#2
Have you been taught the SUVAT equations in GCSE? Such as s = ut + (at^2)/2
0
9 months ago
#3
First, you need to pick the right equation to solve the problem.
In this case, the equation would be s = ut + 0.5*a*t*t (t squared).
(I believe this equation is given in the exam, so no need to memorise it!)
Where:
S = distance travelled ( in this case, 3m)
U = initial velocity ( in this case 0 as the ball was at rest)
T = time taken to travel ( what we are aiming to get)
And a = acceleration. Acceleration of free fall is 9.81m/s/s, or 10m/s/s to be simple.
Hence:
3 = 0t + (0.5*10*t*t)
3 = 0 + (5*t*t)
3 = 5*t*t
3/5 = t*t
Square root (0.6) = t
T = 0.775 seconds ( to1.d.p)
1
Thread starter 9 months ago
#4
(Original post by Doctor1234)
Have you been taught the SUVAT equations in GCSE? Such as s = ut + (at^2)/2
is it the same concept to this equation: v^2 = u^2 + 2as
0
Thread starter 9 months ago
#5
(Original post by Tu quoque)
First, you need to pick the right equation to solve the problem.
In this case, the equation would be s = ut + 0.5*a*t*t (t squared).
(I believe this equation is given in the exam, so no need to memorise it!)
Where:
S = distance travelled ( in this case, 3m)
U = initial velocity ( in this case 0 as the ball was at rest)
T = time taken to travel ( what we are aiming to get)
And a = acceleration. Acceleration of free fall is 9.81m/s/s, or 10m/s/s to be simple.
Hence:
3 = 0t + (0.5*10*t*t)
3 = 0 + (5*t*t)
3 = 5*t*t
3/5 = t*t
Square root (0.6) = t
T = 0.775 seconds ( to1.d.p)
thank you for this detailed explanation, i better consolidate it!
0
9 months ago
#6
(Original post by giolaw)
is it the same concept to this equation: v^2 = u^2 + 2as
All of the SUVAT equations are used for motion calculations, but you need to think about the variables involved. In the equation you stated here you need s, u, v and a. You know a = 9.81 do to gravity, s = 3.0 is given and we assume that u = 0 as it's starting from rest, but we don't know the value of v, making this equation the wrong one to use.
0
9 months ago
#7
@giolaw
No problem!
I suggest you get a grip of the equations first before you start doing past papers, and have the ones that you will be given in the exam in front of you - it would just make things easier.
In case you are wondering, here is the link towards equations that you need to know and the ones you will be given:

Ones that will be given in the exam:
https://filestore.aqa.org.uk/resourc...463-INS-ES.PDF

Ones that you need to memorise:
http://www.rswebsites.co.uk/science/...Simplified.pdf
0
9 months ago
#8
(Original post by ProperLad)
All of the SUVAT equations are used for motion calculations, ....
All SUVAT equations are only valid for motion with constant acceleration. Problems that involve non-constant acceleration motion cannot be solved using SUVAT equations.
1
9 months ago
#9
(Original post by giolaw)
is it the same concept to this equation: v^2 = u^2 + 2as
It's part of the same group but you need the one with t (time) in it as the question asks how long?
0
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