Bomber attack!

Classical Mechanics Level 2

A bomber aircraft is flying at an altitude of 170 m \SI{170}{\meter} at an angle of 3 0 ∘ 30^\circ to the horizontal, as shown in the diagram. At time zero, it releases a bomb aimed at its target. After the release, how much time (in seconds) will the bomb take to hit the target? Give your answer to 2 decimal places.

Details and assumptions:

  • g = 10 m/s 2 g= 10 ~ \text{m/s}^2
  • v 0 = 40 m/s v_0 = 40 ~ \text{m/s}
  • Neglect air resistance.


The answer is 4.16.

Munem Shahriar by Munem Shahriar , 18, Bangladesh

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1 solution

Arjen Vreugdenhil
Nov 29, 2017

In projectile motion we have y = y 0 + v y , 0 t + 1 2 g t 2 . y = y_0 + v_{y,0}t + \tfrac12gt^2. Substitute what we know: y 0 = 170 y_0 = 170 ; v y , 0 = − 40 sin ⁡ 3 0 ∘ = − 20 v_{y,0} = -40\sin 30^\circ = -20 ; g = − 10 g = -10 . Then y = 170 − 20 t − 5 t 2 . y = 170 - 20t - 5t^2. Impact happens when y = 0 y = 0 . We solve for t t . 170 − 20 t − 5 t 2 = 0 170 - 20t - 5t^2 = 0 t 2 + 4 t − 34 = 0 t^2 + 4t - 34 = 0 ( t + 2 ) 2 = 38 (t + 2)^2 = 38 t = − 2 ± 38 . t = -2\pm\sqrt{38}. We need the positive solution: t = − 2 + 38 ≈ 4.16. t = -2 + \sqrt{38} \approx 4.16.

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