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Basic physics logic help - kinematic equations and forces

  • Thread starter FaraDazed
  • Start date
  • #1
347
2

Homework Statement


A small object, of mass 0.02Kg is at the top of a tall structure 160m high, it is dropped from rest.

Ignoring air resistance..

1. Find the speed of the object as it hits the ground
2. find the time taken



Homework Equations


So far we have learned the equations of kinematic motion

v=u+at
s=1/2(u+v)t
2s=(2u+at)t
v^2=u^2+2as

and forces such as

F=ma
Ns=mv-mu



The Attempt at a Solution



I get and can do the math(I think) on this particular question I am a bit puzzled as to some of the phrases and what the variables would be and then what equations and resolving thereof to use.

The obvious ones are "s=160" "u=0" and "m=0.02". Im assuming that part 1 is asking for the final velocity (v) but cant figure out where to start.

If the acceleration was known then I could work out what the final velocity was, however how can I find the acceleration from just that?

Part 2 would be easy if I knew what either the acceleration or final velocity are.

Any help appreciated....I'm sure I'm just reading it wrong or something stupid like making it more complicated than it needs to be.
 

Answers and Replies

  • #2
lewando
Homework Helper
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If the tall structure is on planet Earth, then a = 9.81m/s2 (pointing downward). You do not need to determine this, it is a "given".
 
  • #3
Ibix
Science Advisor
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What did Galileo discover about things free-falling under gravity?
 
  • #4
347
2
Thanks Lewando + ibix

I cant believe how I stupid I was, of course it is. Please forgive my ignorance I have not had any education outside of british GCSE system which was 7 years ago and have just been thrusted into a A-Level equivalent physics class and its going at a very fast pace.

Our tutur has given g to be 9.8 for this class, so on that basis... "u" could be found as below(if i am correct)

v^2=u^2+2as
v^2=0+(2x9.8x160)
v^2=3136
√3136=56

so part 1 is 56m/s ?
 
Last edited:
  • #5
lewando
Homework Helper
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Repair your units and you will be correct.
 
  • #6
347
2
Part 2 would be the following then (if I am correct)

2s=(2u+at)t
320=(0+9.8xt)t
t^2=320/9.8
t^2=32.65
√32.65=5.71

t=5.71s?
 
  • #7
347
2
Repair your units and you will be correct.
Thanks, I think I am getting the hang of it.

The math/algebra is not as hard as the logic bits for me. Im a little dyslexic so sometimes I can read and re-read the question time and time again and not get what its actually asking of me.
 
  • #8
lewando
Homework Helper
Gold Member
1,349
135
Part 2 would be the following then (if I am correct)

2s=(2u+at)t
320=(0+9.8xt)t
t^2=320/9.8
t^2=32.65
√32.65=5.71

t=5.71s?
Yes. Or since you already determined v, you could use v=u+at. With u being 0, its a faster way.
 

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