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Motion in A Plane

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Physics Part I

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Physics

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
A body is projected with velocity u from the top of building h meter high. Find the angle of projection so that it falls at farthest point.

Given, a body is projected with velocity u. 
Height of the building = h

To find - angle of projection theta = ? 

Let the stone be projected at angle θ to make it fall at farthest point R from the foot of tower. 

              

The equation of trajectory of body is given by, 

straight y space equals straight h space plus space xtan space straight theta space minus space fraction numerator straight g over denominator 2 space straight u squared cos squared straight theta end fraction straight x squared space

space space space space equals space straight h space plus space left parenthesis tan space straight theta right parenthesis straight x space minus space space open parentheses fraction numerator straight g space sec squared straight theta over denominator 2 space straight u squared end fraction close parentheses straight x squared space space space space space space... left parenthesis 1 right parenthesis space

Here comma space left parenthesis 0 comma straight R right parenthesis space lies space on space the space trajectory comma space we space have space therefore

0 space equals space straight h space plus space left parenthesis tan space straight theta right parenthesis straight R space minus space open parentheses fraction numerator straight g space sec squared straight theta over denominator 2 space straight u squared end fraction close parentheses straight R squared space space space space space space space space... left parenthesis 2 right parenthesis
For space straight R space to space be space maximum comma space dR over dθ equals 0.
So comma space differentiating space eq. space left parenthesis 2 right parenthesis space straight w. straight r. straight t space straight theta comma space we space get

0 space equals space 0 space plus space left parenthesis sec squared straight theta right parenthesis straight R space plus space left parenthesis tan space straight theta right parenthesis dR over dθ
space space space space equals space minus fraction numerator straight g over denominator 2 straight u squared end fraction open parentheses sec squared straight theta close parentheses 2 straight R dR over dθ space minus space fraction numerator straight g over denominator 2 straight u squared end fraction open parentheses 2 sec squared straight theta space tanθ close parentheses straight R squared

On space putting space dR over dθ space equals space 0 comma space we space get
space space space space space space space space space space left parenthesis sec squared straight theta right parenthesis straight R space minus space fraction numerator straight g over denominator 2 straight u squared end fraction open parentheses 2 sec squared straight theta space tanθ close parentheses straight R squared space equals space 0
straight i. straight e. comma space space space space space space tanθ space equals space straight u squared over gR space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis

Now comma space substituting space value space of space tan space straight theta space in space equation space left parenthesis 2 right parenthesis comma space we space get

space space space space space space space space straight h space plus straight u squared over gR straight R space minus space open square brackets fraction numerator straight g over denominator 2 straight u squared end fraction open parentheses 1 plus fraction numerator straight u to the power of 4 over denominator straight g squared straight R squared end fraction close parentheses close square brackets straight R squared space equals space 0 space

rightwards double arrow space space space space straight h space plus space straight u squared over straight g space minus fraction numerator straight g over denominator 2 straight u squared end fraction straight R squared space minus space 1 half straight u squared over straight g space equals space 0

rightwards double arrow space space space space space space straight u squared space plus space 2 gh space equals space fraction numerator straight g squared straight R squared over denominator straight u squared end fraction equals open parentheses gR over straight u squared close parentheses squared straight u squared space

rightwards double arrow space space space space space space space space straight u squared over gR space equals space fraction numerator straight u over denominator square root of straight u squared plus 2 gh end root end fraction space space space space space space space space space space space space... left parenthesis 4 right parenthesis

Now comma space from space equations space 3 right parenthesis space and space left parenthesis 4 right parenthesis comma space we space have space

tanθ space equals space fraction numerator straight u over denominator square root of straight u squared plus 2 gh end root end fraction
That space is comma space

straight theta space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator straight u over denominator square root of straight u squared plus 2 gh end root end fraction close parentheses

straight theta is the required angle og projection so that the body falls at the farthest point. 






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