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Physics Part I

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A body is projected with velocity u at angle θ with vertical. At what height will it strike the wall, r distance away from the point of projection?

Consider a body which is projected from O in XY plane with velocity 'u' at angle θ with vertical, and
The body is projected at an angle α = 90°–θ with the horizontal. 

Let the body strike the wall at height h.

So, co-ordinates of point P where the projectile hits the wall are ↔ (r,h).

Equation of trajectory is given by, 

space space space space straight y equals xtanα minus 1 half straight g fraction numerator straight x squared over denominator straight u squared cos squared straight alpha end fraction 

     equals xcotθ minus 1 half straight g. fraction numerator straight x squared over denominator straight u squared sin squared straight theta end fraction        [∵ straight alpha equals 90 degree minus straight theta

Since straight P left right arrow left parenthesis straight r comma space straight h right parenthesis lies on trajectory, 

∴        straight h equals rcotθ minus 1 half straight g fraction numerator straight r squared over denominator straight u squared sin squared straight theta end fraction

is the height at which the body will strike the wall.


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