Question

guys :-

solution required:-

40 Qno:-
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Answer 1

Where is the question

Answer 2

Given that the particle covers the height h = 45 m in time t = 1 sec.

Therefore,

h=ut+12gt2⇒45=u×1+5×1×1⇒u=40 ms/

So, this initial velocity is gained by travelling from rest. Therefore, let the distance of the point from where the particle is starting is h' height above the building. Therefore

v=u+at⇒40=0+g×t⇒t=4 sec.

∴ h'=v2−u22a=402−002×10=160020⇒h'=80 m

So, the height of the building is,

H=h+h'=45+80 ⇒H=125 m

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