A bomber plane moving horizontally with a speed of $720 k m / h o u r$ , releases a bomb. The bomb strikes the ground in $10 s$ . What is the angle at which the bomb strikes the ground? $(g = 10 m / s^{2})$

{20.5}^{\circ}

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{26.5}^{\circ}

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{30.5}^{\circ}

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{40.5}^{\circ}

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Solution

The correct option is D ${40.5}^{\circ}$
A bomber plane moving horizontally with a speed $v = 720 k m / h = 85.40 m / s$
Time internal $t = 10 s$
$g = 10 m / s^{2}$
$v = u + a t v = 0 + 10 \times 10 = 100 m / s$
In $Δ A B D$
$tan θ = \frac{A D}{B D} = \frac{85.4}{100} = 0.85 θ = t a n^{- 1} (0.85) = 40.5 °$

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