# F.4 Phy mechanics

RE-EX2

MC3.A rock is released from the second floor an it strikes the ground at a speed of v.If it is released from the seventh floor,what is the speed (in terms of v) when it strikes the groung?(the building:G/F-8/F)

A.v

B.1.87v

C.1.73v

D.3.5v

### 2 Answers

- 6 years agoFavorite Answer
By the formula v^2= u^2 +2as,

v^2 = 2as (Since the initial speed of the rock is zero)

v= sqrt( 2as ) (sqrt =square root )

Let the height of the second floor = d , the seventh floor = 3.5d

Consider when the rock is released from the second floor:

v = sqrt( 2as )

v = sqrt( 2ad ) ------ *

Consider when the rock is released from the seventh floor:

v" = sqrt( 2as )

v" = sqrt( 2a(3.5d) )

v" = sqrt(3.5) * sqrt( 2ad)

Therefore, v" = sqrt (3.5) v (by * )

v" = 1.87 v

The answer is B

Please correct me if I have made any mistakes, or there is a faster way to solve the question ^_^ !

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