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1. An object is dropped from a cliff. How far does the object fall in the third second? Assume that g=10ms^-2

A) 10 metres
B) 20 metres
C) 25 metres
D) 45 metres

Please help, I have been stuck on this question for days!! Any help would be much appreciated.
The answer I keep getting is 45 metres but the mark scheme says something else
2. Well if after every second, it is travelling 10ms^-1 faster than the second before, and that the third second refers to second two -> second three (as in t=2 -> t=3)
at t=2 it is travelling at 20ms^-1, and at t=3 it is travelling at 30ms^-1, from that we can find an average velocity over the third second and hence the distance travelled.
3. I got 25 metres.
We know the variables u (initial velocity) = 0ms^-1 and acceleration = 10ms^-2.
To find the distance travelled in the 3rd second you need to subtract the distance travelled in the first 2 seconds from the distance travelled in 3 seconds.

Using the SUVAT equation S = ut + (1/2)at^2, where u = 0 ms^-1:

distance travelled in the 3rd second = (1/2)*10*3^2 - (1/2)*10*2^2 = 25m.

Hope this helps
(If you see any mistakes please let me know )
4. (Original post by Jas1947)
An object is dropped from a cliff. How far does the object fall in the third second? Assume that g=10ms^-2

A) 10 metres
B) 20 metres
C) 25 metres
D) 45 metres

Please help, I have been stuck on this question for days!! Any help would be much appreciated.
The answer I keep getting is 45 metres but the mark scheme says something else
Just to check, does the mark scheme say that 25m is the answer?
Too late, someone got there before me sorry, page was taking ages to load, but if you still need an explanation, don't hesitate to ask :-)
5. (Original post by Pedals)
I got 25 metres.
We know the variables u (initial velocity) = 0ms^-1 and acceleration = 10ms^-2.
To find the distance travelled in the 3rd second you need to subtract the distance travelled in the first 2 seconds from the distance travelled in 3 seconds.

Using the SUVAT equation S = ut + (1/2)at^2, where u = 0 ms^-1:

distance travelled in the 3rd second = (1/2)*10*3^2 - (1/2)*10*2^2 = 25m.

Hope this helps
(If you see any mistakes please let me know )
Thank you so much! Oh it seems so simple now, i feel stupid being stuck on this for days , thank you so much !
6. (Original post by sue99)
Just to check, does the mark scheme say that 25m is the answer?
Too late, someone got there before me sorry, page was taking ages to load, but if you still need an explanation, don't hesitate to ask :-)
Yes the mark scheme says 25 metres

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