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Q.

At what angle should a body be projected with a velocity 24 ms^-1 just to pass over the obstacle 14 m high at a distance of 24 m? [Take, g =10 ms^-2]

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a

$\tan^{- 1} (3.8)$

b

$\tan^{- 1} (1)$

c

$\tan^{- 1} (3.2)$

d

$\tan^{- 1} (2)$

answer is A, B.

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Detailed Solution

$\begin{array}{ll} y = xtan θ - \frac{{gx}^{2}}{2 u^{2} \cos^{2} θ} \\ \Rightarrow & 14 = 24 \tan θ - \frac{5 (24)^{2}}{(24)^{2}} \sec^{2} θ \\ \Rightarrow & 5 \tan^{2} θ - 24 \tan θ + 19 = 0 \\ \Rightarrow & \tan θ = 1 and \tan θ = \frac{19}{5} = 3 .8 \\ \Rightarrow & θ = \tan^{- 1} (1) and θ = \tan^{- 1} (3 .8) \end{array}$

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