A berry is hanging on a tree at a height 13m above ground and to hit it a boy standing 50m from the tree throws a stone at 1010 m/s. If he wants to hit the berry in minimum time. Find the angle of projection

u=1010m/sEqn of trajectory of stone y=xtanθ−gx22u2cos2θ25tan2θ−100tanθ+51=0

A berry is hanging on a tree at a height 13m above ground and to hit it a boy standing 50m from the tree throws a stone at 1010 m/s. If he wants to hit the berry in minimum time. Find the angle of projection

u=1010m/sEqn of trajectory of stone y=xtanθ−gx22u2cos2θ25tan2θ−100tanθ+51=0

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A berry is hanging on a tree at a height 13m above ground and to hit it a boy standing 50m from the tree throws a stone at $10 \sqrt{10} m / s$ . If he wants to hit the berry in minimum time. Find the angle of projection

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$\tan^{- 1} (\frac{5}{3})$

$\tan^{- 1} (\frac{17}{5})$

$\tan^{- 1} (\frac{3}{5})$

$\tan^{- 1} (\frac{5}{17})$

answer is C.

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Detailed Solution

$u = 10 \sqrt{10} m / s$

Eqn of trajectory of stone $y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ}$

$25 \tan^{2} θ - 100 \tan θ + 51 = 0$

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A berry is hanging on a tree at a height 13m above ground and to hit it a boy standing 50m from the tree throws a stone at 1010 m/s. If he wants to hit the berry in minimum time. Find the angle of projection

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A berry is hanging on a tree at a height 13m above ground and to hit it a boy standing 50m from the tree throws a stone at $10 \sqrt{10} m / s$ . If he wants to hit the berry in minimum time. Find the angle of projection