A projectile is fixed at an angle $θ$ with the horizontal, (as shown in the figure), condition under which it lands perpendicular on an inclined plane of inclination $α$ is

sin α = cos (θ - α)

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cos α = sin (θ - α)

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2 tan α = cot (θ - α)

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cot (θ - α) = sin α

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Solution

The correct option is C $2 tan α = cot (θ - α)$
The time is given as,

$v = u + a t$

$0 = v cos (θ - α) - (g sin α) t$

$t = \frac{v cos (θ - α)}{(g sin α)}$

The time of flight is given as,

$T = \frac{2 v sin (θ - α)}{g cos α}$

Since, both times are equal so, it can be written as,

$\frac{2 v sin (θ - α)}{g cos α} = \frac{v cos (θ - α)}{(g sin α)}$

$cot (θ - α) = 2 tan α$

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A projectile is fixed at an angleθ with the horizontal, (as shown in the figure), condition under which it lands perpendicular on an inclined plane of inclination α is

The correct option is C 2tanα=cot(θ−α)The time is given as, v=u+at 0=vcos(θ−α)−(gsinα)t t=vcos(θ−α)(gsinα) The time of flight is given as, T=2vsin(θ−α)gcosα Since, both times are equal so, it can be written as, 2vsin(θ−α)gcosα=vcos(θ−α)(gsinα) cot(θ−α)=2tanα

A projectile is fixed at an angle $θ$ with the horizontal, (as shown in the figure), condition under which it lands perpendicular on an inclined plane of inclination $α$ is