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A bomber flying with a horizontal velocity of 500 km/h at an altitude of 5 km wants to hit a train moving with a constant velocity of 100 km/h in the same direction and in the same vertical plane. The angle θ between the line of sight of the target and the horizontal at the instant the bomb should be released is

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$55^{\circ}$

$57^{\circ}$

$72^{\circ}$

$27^{\circ}$

answer is A.

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Detailed Solution

Velocity of train relative to bomber is u_r, then

$\begin{array}{l} u_{r} = 500 - 100 = 400 km / h \\ = 400 \times (5 / 18) = (1000 / 9) m / s \end{array}$

The situation is shown fig. (8).

Since $h = \frac{{gx}^{2}}{2 u_{r}^{2}}$

$\begin{array}{l} ∴ 5000 = \frac{{gx}^{2}}{2 (1000 / 9)^{2}} = \frac{10 x^{2}}{2 \times (1000 / 9)^{2}} \\ x = 3513 .65 m \end{array}$

Further, $\tan θ = \frac{h}{x}$

or $\tan θ = \frac{5000}{3513 \cdot 65} = 1 \cdot 423$

or $θ = \tan^{- 1} (1 \cdot 423) = 55^{\circ}$

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