Question 122: Motion in 2D & Vectors

Question

A plane is inclined at an angle $\alpha $ = 30° with respect to the horizontal. A particle is projected with a speed u =
2 ms
–1
, from the base of the plane, making an angle $	heta $ = 15° with respect to the plane as shown in the figure.
the distance from the base, at which the particle hits the plane is close to :
(Take g = 10 ms
–2
)
![](https://imagex.cdn.examgoal.net/7kejek19o8u0w/7ffb6326-f355-4ed0-a19c-0f3135275135/ea4ea000-e54b-11e9-a7bc-b7b08ce8d541/file-7kejek19o8u0x-336w.jpg)

Accepted Answer

$T = {{2u\sin 	heta } \over {g\cos \alpha }}$
$R = u\cos 	heta T - {1 \over 2}g\sin \alpha {T^2}$
$ = {{u\cos 	heta 2u\sin 	heta } \over {g\cos \alpha }} - {{g\sin \alpha } \over 2}{{4{u^2}{{\sin }^2}	heta } \over {{g^2}{{\cos }^2}\alpha }}$
$ = {{{u^2}{{\sin }^2}	heta } \over {g\cos \alpha }} - {{{u^2}\sin \alpha } \over {g{{\cos }^2}\alpha }}\left\{ {1 - \cos 2	heta } ight\}$
$ = {{4 	imes {1 \over 2}} \over {10 	imes {{\sqrt 3 } \over 2}}} - {{{u^2}\sin \alpha } \over {g{{\cos }^2}\alpha }}\left\{ {1 - {{\sqrt 3 } \over 2}} ight\}$
$ = {4 \over {10\sqrt 3 }} - {8 \over {30}}\left\{ {1 - {{\sqrt 3 } \over 2}} ight\}$
$ = {4 \over {5\sqrt 3 }} - {8 \over {30}} = {{8\sqrt 3  - 8} \over {30}}$
$ = {{8\left( {\sqrt 3  - 1} ight)} \over {30}} = 20\,cm$

Answer

14 cm

Answer

18 cm

Answer

20 cm

Answer

26 cm

Motion in 2D & Vectors

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