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CBSE Class 11-science Answered

A stone is thrown upwards vertically. It has speed 10m/s when it reaches one half of its maximum height. Then find out how high it will rise?
Asked by udayanmazumder079 | 13 Jul, 2023, 09:21: PM
answered-by-expert Expert Answer
Maximum height hmax reached by the stone is

 begin mathsize 14px style h subscript m a x end subscript space equals space fraction numerator u squared over denominator 2 space g end fraction end style  ...................... (1)
where u is initial speed when the stone is thrown. 
 
At half of maximum height , speed is 10 m/s . 
 
Let us consider the equation of motion " v2 = u2 - 2 g h " , where u is initial speed and
v is speed after making the vertical displacement h .
 
If we substitute v = 10 m/s and h = (1/2)hmax in above equation of motion, then we get
 
begin mathsize 14px style 100 space equals space u squared space minus space left parenthesis space 2 space cross times g space cross times space h subscript m a x subscript space end subscript over 2 right parenthesis end style  .....................(2) 
If we substitute hmax using eqn.(1) , above eqn.(2) becomes 
 
begin mathsize 14px style 100 space equals space u squared space minus space left parenthesis space 2 space cross times g space cross times fraction numerator u squared over denominator 4 space g end fraction right parenthesis space equals space u squared over 2 end style 
From above expression, we get u2 = 200 m2 /s2 
 
Hence , we get maximum height from eqn.(1) as
 
begin mathsize 14px style h subscript m a x end subscript space equals fraction numerator 200 over denominator 2 space cross times space 9.8 end fraction space equals space 10.2 space m end style
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