Physics, asked by ArpitAnand2766, 1 month ago

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 anish28908
31

Answer:

b) 135 m

Explanation:

believe me it is right option

Answered by singhranjeet6856
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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