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Question Number 16044 by Tinkutara last updated on 17/Jun/17
A particle is projected horizontally with speed u from point A, which is 10 m above the ground. If the particle hits the inclined plane perpendicularly at point B. [g = 10 m/s^2 ] Find horizontal speed with which the particle was projected.
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{horizontally} \\ $$$$\mathrm{with}\:\mathrm{speed}\:{u}\:\mathrm{from}\:\mathrm{point}\:{A},\:\mathrm{which}\:\mathrm{is}\:\mathrm{10} \\ $$$$\mathrm{m}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{hits} \\ $$$$\mathrm{the}\:\mathrm{inclined}\:\mathrm{plane}\:\mathrm{perpendicularly}\:\mathrm{at} \\ $$$$\mathrm{point}\:{B}.\:\left[{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right] \\ $$$$\mathrm{Find}\:\mathrm{horizontal}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{the} \\ $$$$\mathrm{particle}\:\mathrm{was}\:\mathrm{projected}. \\ $$
Commented by Tinkutara last updated on 17/Jun/17
Answered by ajfour last updated on 17/Jun/17
As the projectile hits a 45° plane normally, its velocity there (at B) is directed at an angle of 45° below the horizontal. (v_y )_(at B) =−(v_x )_(at B) −gt = −u ⇒ t=(u/g) also if origin is taken at the the start of incline, at point B, y=x h−(1/2)gt^2 = ut , with t=u/g we have h−(g/2)((u^2 /g^2 ))=(u^2 /g) h= (u^2 /(2g))+(u^2 /g) ⇒ ((3u^2 )/(2g)) = h or u = (√((2gh)/3)) = (√((2×10×10)/3)) m/s =10(√(2/3)) m/s .
$${As}\:{the}\:{projectile}\:{hits}\:{a}\:\mathrm{45}°\:{plane} \\ $$$${normally},\:{its}\:{velocity}\:{there}\:\left({at}\:{B}\right) \\ $$$${is}\:{directed}\:{at}\:{an}\:{angle}\:{of}\:\mathrm{45}°\: \\ $$$${below}\:{the}\:{horizontal}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({v}_{{y}} \right)_{{at}\:{B}} =−\left({v}_{{x}} \right)_{{at}\:{B}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{gt}\:=\:−{u}\:\:\:\Rightarrow\:\:{t}=\frac{{u}}{{g}} \\ $$$$\:\:\:\:\:\:\:\:\:{also}\:{if}\:{origin}\:\:{is}\:{taken}\:{at}\:{the} \\ $$$$\:{the}\:{start}\:{of}\:{incline},\:\: \\ $$$$\:\:\:\:\:\:\:\:{at}\:{point}\:{B},\:\:\:\:\:{y}={x} \\ $$$$\:\:\:\:\:\:{h}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \:=\:{ut}\:\:,\:\:\:\:{with}\:{t}={u}/{g} \\ $$$${we}\:{have}\:\:\: \\ $$$$\:\:\:\:\:\:\:{h}−\frac{{g}}{\mathrm{2}}\left(\frac{{u}^{\mathrm{2}} }{{g}^{\mathrm{2}} }\right)=\frac{{u}^{\mathrm{2}} }{{g}} \\ $$$$\:\:\:\:\:\:\:\:{h}=\:\frac{{u}^{\mathrm{2}} }{\mathrm{2}{g}}+\frac{{u}^{\mathrm{2}} }{{g}} \\ $$$$\:\:\Rightarrow\:\:\:\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}{g}}\:=\:{h} \\ $$$$\:\:{or}\:\:{u}\:=\:\sqrt{\frac{\mathrm{2}{gh}}{\mathrm{3}}}\:=\:\sqrt{\frac{\mathrm{2}×\mathrm{10}×\mathrm{10}}{\mathrm{3}}}\:{m}/{s} \\ $$$$\:\:\:\:\:\:=\mathrm{10}\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}\:{m}/{s}\:. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Tinkutara last updated on 17/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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