Question

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A Shell is fired from a gun from the bottom of a hill along its slope.
the slope of the hill is alpha equals to 30 degree and the angle of The parallel to the horizontal beta equals to 60 degree the initial velocity u of the cell is 21 second inverse find the distance from the gun to the point at which the shell Falls
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Answer 1

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Refer To The Attachment and your answer is 30

Answer 2

The horizontal and the vertical distances covered by the shell in time t are given by

$x = u \cos \beta .t$

and

$y = u \sin \beta .t - \frac{1}{2} gt ^{2}$

If the shell falls on the incline plane at distance OP=s, then

$x = s \cos \alpha$

$y = s \sin\alpha$

From (1)and (3),

$s \cos\alpha = u \sin \beta.t$

or

$t = \frac{s \cos\alpha }{u \cos\beta }$

From (2)and (4),

$s \sin \alpha = u \sin \beta.t - \frac{1}{2} gt^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = u \: \sin\beta . \frac{s \cos( \alpha ) }{u \cos( \beta ) } - \frac{1}{2} g. \frac{s ^{2} \cos^{2} \alpha }{u ^{2} \cos^{2} \beta }$

There fore,

$s = \frac{2u^{2} }{g} ( \frac{ \sin \beta \cos \alpha - \cos \beta \sin \alpha }{ { \cos }^{2} \alpha } ) \cos( \beta ) \\ \: \: \: \: = \frac{2u ^{2} \sin( \beta - \alpha ) \cos( \beta ) }{g \cos ^{2} \alpha } \\ \: \: \: \: = \frac{2 \times (21)^{2} \sin(60 - 30) \cos(60) }{9.8 \times \cos ^{2} 30 } \\ \: \: \: \: = 30m$