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A particle is projected in air at an angle $\beta$ to a surface which itself is inclined at an angle $\alpha$ to the horizontal.

a) find an expression of range on the plane surface

b) time of flight

c) $\beta$ at which range will be maximum

Answers (1)

a) expression of range on the plane surface

Now, $a_{y}=-g\; \cos \alpha$ and $a_{x}= g\; \sin \alpha$

$Y=0$ at O and P,

So

$U_{y}=v\; \sin \beta$ where t = T

We calculate Time of Flight Part (b) before Part (a)

b) motion of a particle along Y axis

$s=ut+\frac{1}{2}gt^{2}$

$s=0,u=v\; \sin \beta , g=a=-g\; \cos \alpha ,t=T$

$0=v\; \sin \beta .T+\frac{1}{2}(-g\; \cos \alpha )T^{2}$

$T[v\; \sin \beta -T.\frac{g}{2}\cos \alpha ]=0$

$\frac{Tg}{2}.\cos \alpha =v\sin \beta$

So, $T=2v\; \sin \beta /g.\cos \alpha$

a) Now we continue the part a

$T=2v\; \sin \beta /g.\cos \alpha$

$s=ut+\frac{1}{2}gt^{2}$

$L=v\; \cos \beta (T)+\frac{1}{2}(-g\; \sin \alpha )T^{2}$

$L=2\; v^{2}\; \sin \beta[\cos \beta \cos \alpha -\sin \beta \sin \alpha ]/g \cos ^{2}\alpha$

$L=2\; v^{2}\; \sin \beta \times \cos [\alpha +\beta ]/g \cos ^{2}\alpha$

c) on the axis X, L will be maximum when $\sin \beta \cos [\alpha +\beta ]$ will be maximum

let $z=\sin \beta \cos [\alpha +\beta ]$

$=\sin \beta [\cos \beta \cos \alpha -\sin \beta \sin \alpha ]$

$=\frac{1}{2}[\cos \alpha\; 2\sin\; \beta +\sin \alpha \cos \; 2\; \beta-\sin \alpha ]$

$=\frac{1}{2}[\sin(2\beta +\alpha )-\sin\; \alpha ]$

In order to make Z maximum, we put $[\sin(2\beta +\alpha )-\sin\; \alpha ]=1$

Opening the brackets, we get $(2\beta +\alpha )=90$

$2\beta =90-\alpha$

$\beta =45-\frac{\alpha }{2}$ radian

Posted by

infoexpert23

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