An inclined plane makes as angle -0 - 30- with the horizontal- A particle is projected from this plane with a speed of 5 m-s at an angle of elevation - - 30- with the horizontal as shown in the figure-Find the range of the particle on the plane when it strikes the plane- Assume the incline is long enough and g-10 m-s2

Angle of inclination with nbsp;the inclined surface, α = θ_0+β = 60^∘ Range on incline, R = u^2g cos^2 θ_0 [sin (2 α θ_0)+ sinθ_0] = 5^210 cos^2 30^∘ [sin ( ...

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Standard XI

Physics

Range on an Incline

An inclined p...

Question

An inclined plane makes as angle $θ_{0} = 30^{\circ}$ with the horizontal. A particle is projected from this plane with a speed of $5 m/s$ at an angle of elevation $β = 30^{\circ}$ with the horizontal as shown in the figure.

Find the range of the particle on the plane when it strikes the plane. (Assume the incline is long enough and $g = 10 {m/s}^{2}$ )

3.5 m

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4 m

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4.2 m

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5 m

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Solution

The correct option is D $5 m$
Angle of inclination with the inclined surface, $α = θ_{0} + β = 60^{\circ}$
Range on incline, $R = \frac{u^{2}}{g {cos}^{2} θ_{0}} [sin (2 α - θ_{0}) + sin θ_{0}]$
$= \frac{5^{2}}{10 {cos}^{2} 30^{\circ}} [sin (2 \times 60^{\circ} - 30^{\circ}) + sin 30^{\circ}]$
$= \frac{25}{10 \times {(\frac{\sqrt{3}}{2})}^{2}} [sin 90^{\circ} + \frac{1}{2}]$
$= \frac{25 \times 4}{10 \times 3} \times \frac{3}{2}$
= $5 m$

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The correct option is D 5 mAngle of inclination with the inclined surface, α=θ0+β=60∘ Range on incline, R=u2gcos2θ0[sin(2α−θ0)+sinθ0] =5210cos230∘[sin(2×60∘−30∘)+sin30∘] =2510×(√32)2[sin90∘+12] =25×410×3×32 = 5 m