Question

A plane is inclined at an angle α = 30º with respect to the horizontal. A particle is projected with a speed u = 2 ms⁻¹, from the base of the plant, making an angle θ = 15º 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 ms²)
(A) 18 cm (B) 14 cm
(C) 26 cm (D) 20 cm

Answer 1

Given:

A plane is inclined at an angle α = 30º with respect to the horizontal. A particle is projected with a speed u = 2 ms⁻¹, from the base of the plant, making an angle θ = 15º with respect to the plane as shown in the figure.

To find:

Range of projectile

Calculation:

Let the angle of projectile w.r.t to ground be $\phi$.

$\therefore \phi = \theta + \alpha$

$= > \phi = 15 + 30$

$= > \phi = {45}^{ \circ}$

Now, please remember the general formula for range on inclined plane :

$R = \dfrac{ {u}^{2} }{g { \cos}^{2}( \alpha ) } \bigg \{ \sin(2 \phi - \alpha ) - \sin( \alpha ) \bigg \}$

Putting all the available values:

$= > R = \dfrac{ {2}^{2} }{g { \cos}^{2}( {30}^{ \circ} ) } \bigg \{ \sin( {90}^{ \circ} - {30}^{ \circ} ) - \sin( {30}^{ \circ} ) \bigg \}$

$= > R = \dfrac{ 16}{ 30} \bigg \{ \sin( {60}^{ \circ} ) - \sin( {30}^{ \circ} ) \bigg \}$

$= > R = \dfrac{ 16}{ 30} \bigg \{ \dfrac{ \sqrt{3} }{2} - \dfrac{1}{2} \bigg \}$

$= > R = \dfrac{ 16}{ 30} \bigg \{ \dfrac{ \sqrt{3} - 1 }{2} \bigg \}$

$= > R = \dfrac{ 16}{ 30} \bigg \{ \dfrac{ 0.732}{2} \bigg \}$

$= > R = 0.195 \: m$

$= > R \approx 0.20 \: m$

$= > R \approx 20 \: cm$

So, final answer:

$\boxed{ \large{ \bold{R \approx 20 \: cm}}}$

Answer 2

Explanation:

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