a bomber flying upward at 53deg with vertical release a bomb at at altitude of 800 m. the bomb strike the ground 20sec after its release.find velocity of bomber at release time

Question

anonymous · Answer

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problem illustration...

anonymous · Answer

From Projectile:
$$y = v_o\sin \theta \times t - \frac{1}{2}gt^2$$where  $\theta = 90-53=37$
Just plug-in other given values, you can solve for  $v_o$ which is the velocity of the bomber at the time the bomb is release...