A ball is projected upwards from the foot of a tower. It crosses the top of the tower twice at an interval of 4s and reaches the ground after 12s from the start. The height of the tower is:
A)120 m
B)135 m
C)175 m
D)160 m
Answers
Answered by
31
Answer:
b) 135 m
Explanation:
believe me it is right option
Answered by
0
Let height of tower be h.
As given,
t
BCD
=6sec
and t
ABCDE
=12sec
∴t
AB
=3sec=t
DC
and , ∴t
BC
=3sec=t
CD
So, from relation v=u+at
From A to C,u=U
v=0,a=−g,t=6sec
O=u−gt,t=
g
u
, or u=10×6m/s
u=60m/s
Now, from relation s=ut+
2
1
at
2
from, A to B,s=h,u=60m/s,t=3sec,a=−g
h=60×3−
2
1
×10×9
h=135m
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