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gg asked in 科學及數學其他 - 科學 · 6 years ago

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

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  • 6 years ago
    Favorite 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 ^_^ !

    Source(s): Me
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  • 6 years ago

    A/C i am still thinking

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