A particle is projected from a point O in the horizontal surface OA with speed u and angle of projection - It just grazes the plane BC which makes an angle - with the horizontal- The time taken by the projectile to reach p from the instant of projection is

The correct option is C u sin(θ α)/g cosα From figure, #160; #160; #160; u_x = ucosθ #160; #160; #160; #160; #160; #160; ...

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Byju's Answer

Standard XII

Physics

Problem Solving

A particle is...

Question

A particle is projected from a point $O$ in the horizontal surface $O A$ with speed $u$ and angle of projection $θ$ . It just grazes the plane $B C$ which makes an angle $α$ with the horizontal. The time taken by the projectile to reach $p$ from the instant of projection is

\frac{2 u s i n θ}{g c o s θ}

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\frac{u s i n θ}{g c o s α}

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\frac{u s i n (θ - α)}{g c o s α}

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\frac{2 u s i n (θ - α)}{g c o s α}

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Solution

The correct option is C $\frac{u s i n (θ - α)}{g c o s α}$
From figure, $u_{x} = u c o s θ$ $u_{y} = u s i n θ$
At point P, the particle just grazes the line BC at the instant $t$ i.e $\frac{V_{y}}{V_{x}} = t a n α$
Also as $a_{x} = 0$ $⟹$ $V_{x} = u_{x} = u c o s θ$
$∴$ $V_{y} = V_{x} t a n α = (u c o s θ) t a n α$
y direction : $V_{y} = u_{y} - g t$
$∴$ $u c o s θ t a n α = u s i n θ - g t$

OR $t = \frac{u [s i n θ - t a n α c o s θ]}{g} = \frac{u [s i n θ c o s α - s i n α c o s θ]}{g c o s α}$

$⟹$ $t = \frac{u s i n (θ - α)}{g c o s α}$

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A particle is projected from a point O in the horizontal surface OA with speed u and angle of projection θ. It just grazes the plane BC which makes an angle α with the horizontal. The time taken by the projectile to reach p from the instant of projection is

A particle is projected from a point $O$ in the horizontal surface $O A$ with speed $u$ and angle of projection $θ$ . It just grazes the plane $B C$ which makes an angle $α$ with the horizontal. The time taken by the projectile to reach $p$ from the instant of projection is