A stone is dropped from a 100m high tower. How long does it take to fall? a) the first 50m and b) the second 50m.
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For the first 50 m
U is 0
And v is equal to
V^2 = u ^2 + 2as
v^2 =0 + 2×10×50 = 1000
v= 10√10
Now t = v-u/t = (10√10 -0)/t
Therefore t = 10√10/10 = √10s =3.34s approx.....
Now for the next 50 m
V = 0
U = 10√10
S = ut + 1/2 at^2
50 = 10√10 t + 1/2 10t^2
50 = 10√10 + 5t^2
t^2 +5√10-10 = 0
t =( -5 +- √25 + 40)/2 =( -5 + √65)/2 = (sssin time cannot be 0)
t = (-5 + 8.06)/2 = 3.06/ 2 1.58 s
Thank u★★★
#ckc
U is 0
And v is equal to
V^2 = u ^2 + 2as
v^2 =0 + 2×10×50 = 1000
v= 10√10
Now t = v-u/t = (10√10 -0)/t
Therefore t = 10√10/10 = √10s =3.34s approx.....
Now for the next 50 m
V = 0
U = 10√10
S = ut + 1/2 at^2
50 = 10√10 t + 1/2 10t^2
50 = 10√10 + 5t^2
t^2 +5√10-10 = 0
t =( -5 +- √25 + 40)/2 =( -5 + √65)/2 = (sssin time cannot be 0)
t = (-5 + 8.06)/2 = 3.06/ 2 1.58 s
Thank u★★★
#ckc
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