JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 2)
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 $$\theta $$ = 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)_10th_April_Evening_Slot_en_2_1.png)
(Take g = 10 ms –2)
14 cm
18 cm
20 cm
26 cm
Explanation
$$T = {{2u\sin \theta } \over {g\cos \alpha }}$$
$$R = u\cos \theta T - {1 \over 2}g\sin \alpha {T^2}$$
$$ = {{u\cos \theta 2u\sin \theta } \over {g\cos \alpha }} - {{g\sin \alpha } \over 2}{{4{u^2}{{\sin }^2}\theta } \over {{g^2}{{\cos }^2}\alpha }}$$
$$ = {{{u^2}{{\sin }^2}\theta } \over {g\cos \alpha }} - {{{u^2}\sin \alpha } \over {g{{\cos }^2}\alpha }}\left\{ {1 - \cos 2\theta } \right\}$$
$$ = {{4 \times {1 \over 2}} \over {10 \times {{\sqrt 3 } \over 2}}} - {{{u^2}\sin \alpha } \over {g{{\cos }^2}\alpha }}\left\{ {1 - {{\sqrt 3 } \over 2}} \right\}$$
$$ = {4 \over {10\sqrt 3 }} - {8 \over {30}}\left\{ {1 - {{\sqrt 3 } \over 2}} \right\}$$
$$ = {4 \over {5\sqrt 3 }} - {8 \over {30}} = {{8\sqrt 3 - 8} \over {30}}$$
$$ = {{8\left( {\sqrt 3 - 1} \right)} \over {30}} = 20\,cm$$
$$R = u\cos \theta T - {1 \over 2}g\sin \alpha {T^2}$$
$$ = {{u\cos \theta 2u\sin \theta } \over {g\cos \alpha }} - {{g\sin \alpha } \over 2}{{4{u^2}{{\sin }^2}\theta } \over {{g^2}{{\cos }^2}\alpha }}$$
$$ = {{{u^2}{{\sin }^2}\theta } \over {g\cos \alpha }} - {{{u^2}\sin \alpha } \over {g{{\cos }^2}\alpha }}\left\{ {1 - \cos 2\theta } \right\}$$
$$ = {{4 \times {1 \over 2}} \over {10 \times {{\sqrt 3 } \over 2}}} - {{{u^2}\sin \alpha } \over {g{{\cos }^2}\alpha }}\left\{ {1 - {{\sqrt 3 } \over 2}} \right\}$$
$$ = {4 \over {10\sqrt 3 }} - {8 \over {30}}\left\{ {1 - {{\sqrt 3 } \over 2}} \right\}$$
$$ = {4 \over {5\sqrt 3 }} - {8 \over {30}} = {{8\sqrt 3 - 8} \over {30}}$$
$$ = {{8\left( {\sqrt 3 - 1} \right)} \over {30}} = 20\,cm$$
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