A body drops from 10m, it will hit the ground in
- square_root(5 seconds).
- square_root(4 seconds).
- square_root(2 seconds).
Explanation
For free fall from rest, distance s = half g t squared. Taking g as about 10 m/s squared, 10 = 5 t squared gives t squared = 2, so t = the square root of 2 seconds. The other options do not satisfy the fall equation for a 10 m drop.
Ptarbin asking:
What formula is used for this?
Community Comments (4)
P
Ptarbin Posts: 3 13.01.2016 / 22:35
It's ok I've worked it out now. S = ut+1/2 a t^2
D
dimky Posts: 514 14.01.2016 / 14:14
Yes, exactly. :)
S
stubbsy1875 Posts: 1 24.08.2017 / 08:42
Still think it's wrong though? S=distance.... and that's given in the question.
I thought it should have been 10m ÷ g(9.81m/s) or am I missing something?
I thought it should have been 10m ÷ g(9.81m/s) or am I missing something?
J
JD2016 Posts: 54 27.08.2017 / 14:02
What you are missing is the 2 and the square or square root!
S=distance =10
a= acceleration =g = 9.81 m/s2 - rounded up to 10.
u = initial velocity = 0 (assuming it is dropped from rest at 10m).
Fill in the equation:
10 = 0 + 0.5 x 10t**2
2= t**2
t= square root of 2 seconds = 1.414 seconds.
S=distance =10
a= acceleration =g = 9.81 m/s2 - rounded up to 10.
u = initial velocity = 0 (assuming it is dropped from rest at 10m).
Fill in the equation:
10 = 0 + 0.5 x 10t**2
2= t**2
t= square root of 2 seconds = 1.414 seconds.
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