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Class 10 Class 12

સમતલમાં ગતિ

એક પદાર્થને $stack straight v space with rightwards arrow on top$ જેટલા વેગથી પ્રક્ષિપ્ત કરવામાં આવે છે. જિ તેની અવધિ મહત્તમ ઊંચાઈ કરતાં બમણી હોય, તો અવધિનું મૂલ્ય .........
(સૂચન : ઉપરના પ્રશ્ન – 6 ના પરિણામનો ઉપયોગ કરો.)

$straight v squared over straight g$
$3 over 5 space v squared over g$
$4 over 5 space v squared over g$
$1 half space v squared over g$

જો $stack straight A space with rightwards arrow on top equals 2 i with overparenthesis on top plus 3 j with overparenthesis on top minus k with overparenthesis on top$ અને $stack straight B space with rightwards arrow on top equals 4 i with overparenthesis on top plus 6 j with overparenthesis on top minus 2 k with overparenthesis on top$ હોય, તો $stack straight A space with rightwards arrow on top$ અને $stack straight B space with rightwards arrow on top$ વચ્ચેનો ખૂણો

$straight pi$
$straight pi over 3$
$straight pi over 2$
$0 to the power of degree$

પ્રક્ષિપ્ત ગતિમાં પદાર્થે પ્રાપ્ત કરેલી મહત્તમ ઊંચાઈ તેની અવધિ કરતાં અડધી હોય, $open parentheses straight H equals 1 half straight R close parentheses$ નો સમક્ષિતિજ સાથેનો પ્રક્ષિપ્ત કોણ $straight theta subscript 0$ .........

$tan to the power of negative 1 end exponent open parentheses 1 close parentheses$
$tan to the power of negative 1 end exponent open parentheses 2 close parentheses$
$tan to the power of negative 1 end exponent open parentheses 3 close parentheses$
$tan to the power of negative 1 end exponent open parentheses 4 close parentheses$

tan to the power of negative 1 end exponent open parentheses 2 close parentheses

Tips: -

straight H equals fraction numerator straight v subscript 0 squared space sin squared space straight theta subscript 0 space over denominator space 2 straight g end fraction straight R equals fraction numerator straight v subscript 0 squared space sin space 2 space straight theta subscript 0 over denominator straight g end fraction equals fraction numerator 2 straight v subscript 0 squared space sin space straight theta subscript 0 space cos space straight theta subscript 0 over denominator straight g end fraction

therefore space straight H over straight R equals fraction numerator straight v subscript 0 squared space sin squared space straight theta subscript 0 over denominator 2 straight g end fraction cross times fraction numerator straight g over denominator 2 straight v subscript 0 squared space sin space straight theta subscript 0 space cos space straight theta subscript 0 end fraction space space space space space space space space space space space space space equals space space fraction numerator sin space straight theta subscript 0 over denominator 4 space cos space straight theta subscript 0 end fraction therefore space straight H over straight R space equals fraction numerator tan space straight theta subscript 0 over denominator 4 end fraction therefore space fraction numerator open parentheses begin display style straight R over 2 end style close parentheses over denominator straight R end fraction equals fraction numerator tan space straight theta subscript 0 over denominator 4 end fraction therefore space tan space straight theta subscript 0 equals space 2 space rightwards double arrow space straight theta subscript 0 space equals space tan to the power of negative 1 end exponent space open parentheses 2 close parentheses

એક ભૌતિક રાશિને રજૂ કરતો સદિશ $stack straight C space with rightwards arrow on top equals 2 i with overparenthesis on top plus 3 j with overparenthesis on top plus 4 k with overparenthesis on top$ છે, તો X-અક્ષ અને આ સદિશ વચ્ચેનો કોણ .......

$cos to the power of negative 1 end exponent fraction numerator 3 over denominator square root of 29 end fraction$
$cos to the power of negative 1 end exponent fraction numerator 4 over denominator square root of 29 end fraction$
$cos to the power of negative 1 end exponent fraction numerator 5 over denominator square root of 29 end fraction$
$cos to the power of negative 1 end exponent fraction numerator 2 over denominator square root of 29 end fraction$

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