A stone is dropped from a 100m high tower. How long does it take to fall? a) The first 50m and 2) The second 50m
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Well for the first 50m we have ;
u=-0m/s
a=-10m/s
s=-50m
t=?
Therefore, by 2nd eq of motion we have;
s=ut+1/2at^2
Substitution of the known...
-50=0×t+1/2×-10×t^2
-100/-10=t^2
sqrt (10)=t
Now the time taken to cover next 50m would be = time taken for covering 100m - time taken for covering the first 50m
by 2nd eq of motion
-100=0×T+1/2×-10×T^2
T=sqrt (20)
therefore the time taken to cover next 50m is=
=> sqrt (20)-sqrt (10)
hope it helps.....
u=-0m/s
a=-10m/s
s=-50m
t=?
Therefore, by 2nd eq of motion we have;
s=ut+1/2at^2
Substitution of the known...
-50=0×t+1/2×-10×t^2
-100/-10=t^2
sqrt (10)=t
Now the time taken to cover next 50m would be = time taken for covering 100m - time taken for covering the first 50m
by 2nd eq of motion
-100=0×T+1/2×-10×T^2
T=sqrt (20)
therefore the time taken to cover next 50m is=
=> sqrt (20)-sqrt (10)
hope it helps.....
mrigansh:
pls mark it as best
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