Question

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A particle in air at an angle β to a surface which itself is inclined at an angle α to the horizontal.(see figure)

(a) Find expression for range on the plane surface
(b) Time of flight
(c) β at which range will be maximum

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Answer 1

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$\huge{\bold{\underline{Given:-}}}$

• Angle of inclination = α
• Angle at which Projectile is Projected = β.

$\huge{\bold{\underline{Explanation:-}}}$

Here the Acceleration due to gravity will be acting vertically downwards,

But For the Projectile a Component of acceleration due to gravity will be responsible, I.e. gcosα.

And, The Final displacement in y - direction is Zero .

Now,

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Applying Second kinematic equation in y - direction,

$\large{\boxed{\tt S_y = u_yt + \dfrac{1}{2}a_yt^2}}$

Substituting the values,

$\large{\tt y = (v_o sin \beta)t + \dfrac{1}{2} \times (- g cos \alpha)t^2}$

Here the component of Velocity along the y - direction is taken, and Acceleration is taken as negative because it is opposing the motion of the body.

$\large{\tt 0 = (v_o sin \beta)t - \dfrac{(g cos \alpha)t^2}{2}}$

$\large{\tt (v_o sin \beta)t = \dfrac{(g cos \alpha)t^2}{2}}$

$\large{\tt (v_o sin \beta)\cancel{t} = \dfrac{(g cos \alpha)\cancel{t^2}}{2}}$

$\large{\tt v_o sin \beta = \dfrac{(g cos \alpha)t}{2}}$

For Time of flight; t = T seconds,

$\large{\tt (g cos \alpha)T = 2v_o sin \beta}$

Now,

$\large{\boxed{\boxed{\tt T = \dfrac{2v_o sin \beta}{(g cos \alpha)}}}}$

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By 2nd Kinematic equation in x direction,

$\large{\boxed{\tt s_x = u_xt + \dfrac{1}{2}a_xt^2}}$

$\large{\tt R = v_o sin \beta \times \left( \dfrac{2v_o sin \beta}{(g cos \alpha)} \right) + \dfrac{g sin \alpha}{2} \times \left( \dfrac{2v_o sin \beta}{(g cos \alpha)} \right)^2}$

Now,

$\large{\tt R = 2v_o^2 sin \beta \times \left( \dfrac{ sin \beta}{(g cos \alpha)} \right) + \dfrac{g sin \alpha}{2} \times \left( \dfrac{2v_o sin \beta}{(g cos \alpha)} \right)^2}$

Taking common,

$\large{\tt R = \dfrac{2v_o^2 sin \beta}{(g^2 cos^2 \alpha)} ( cos \beta cos \alpha - sin \alpha sin \beta)}$

Now, using the trigonometric identity,

we get

${\tt ( cos \beta cos \alpha - sin \alpha sin \beta) = cos ( \alpha + \beta)}$

$\large{\tt R = \dfrac{2v_o^2 sin \beta}{(g^2 cos^2 \alpha)} \times cos ( \alpha + \beta)}$

$\large{\boxed{\boxed{\tt R = \dfrac{2v_o^2 sin \beta cos(\alpha + \beta)}{(g^2 cos^2 \alpha)}}}}$

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#refer the attachment for detailed explanation of the third part of the answer.

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