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Home / Doubts and Solutions/JEE/Mathematics

Ques 37

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Asked by rbhatt16 10th July 2018, 11:42 PM
Answered by Expert
Answer:
 
First we find the range of the stone that has projected horizontally. we call this stone first stone
 
let t be the time for this stone to hit the ground. initial vertical componenet of velocity is zero.
 
hence we have, begin mathsize 12px style h space equals space 1 half g space t squared space space space o r space t space equals space square root of fraction numerator 2 space h over denominator g end fraction end root end style 
In this time t, the distance S travelled by the first stone is given by,    S = begin mathsize 12px style u cross times square root of fraction numerator 2 space h over denominator g end fraction end root end style .........................(1)
The height h1 above the cliff that is reached by the stone projected with an angle θ ( we call this stone as second stone)  is given by
 
begin mathsize 12px style h subscript 1 space equals space fraction numerator u squared sin squared theta over denominator 2 g end fraction end style ..................(2)
time t1 to reach height h1 is given by,  t1 = (u sinθ) / g ......................(3)
 
after reaching the maximum height h1, let t2 be the time taken to travel the height (h1+h) in order to hit the ground for the second stone. 
 
begin mathsize 12px style h subscript 1 space plus space h space equals space fraction numerator u squared sin squared theta over denominator 2 g end fraction plus h space equals space 1 half g space t subscript 2 superscript 2 space end style ..................(4)
Now horizontal distance travelled by the second stone during the time (t1+t2) and the distance given in eqn.(1) are same.
 
Hence we have
 
begin mathsize 12px style open parentheses t subscript 1 space plus space t subscript 2 close parentheses cross times u space cos theta space equals space u cross times square root of fraction numerator 2 space h over denominator g end fraction end root end style
begin mathsize 12px style t subscript 1 plus t subscript 2 space equals space s e c theta space square root of fraction numerator 2 space h over denominator g end fraction end root space fraction numerator u space sin theta over denominator g end fraction space plus space t subscript 2 space equals space s e c theta space square root of fraction numerator 2 space h over denominator g end fraction end root t subscript 2 space equals space s e c theta space square root of fraction numerator 2 space h over denominator g end fraction end root space minus space fraction numerator u space sin theta over denominator g end fraction 1 half g space t subscript 2 superscript 2 space equals space h space s e c squared theta space plus fraction numerator u squared sin squared theta over denominator 2 g end fraction minus space u space tan theta space square root of fraction numerator 2 space h over denominator g end fraction end root space.......................... left parenthesis 5 right parenthesis end style
By equating (4) and (5), we have
 
 begin mathsize 12px style h space equals space h space s e c squared theta space minus space u space tan theta space square root of fraction numerator 2 space h over denominator g end fraction end root space.................. left parenthesis 6 right parenthesis end style
Eqn.(6) can be simplified to get ,    begin mathsize 12px style tan space theta space equals space u square root of fraction numerator 2 over denominator g space h end fraction end root end style
 
 
Answered by Expert 11th July 2018, 2:54 PM
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