Hi, It would be wonderful if someone out there could help me figurethis problem

Question

Hi,

It would be wonderful if someone out there could help me figurethis problem out. I figured out part 1 but can not get part2.

You are working as a consultant on a video game designing a bombsite for a World War I airplane. In this game, the plane you areflying is traveling horizontally at 38.6m/s at an altitude of 100 m when it dropsa bomb.

(a) Determine how far horizontally from the target you shouldrelease the bomb. The answer is 174.4 m
(b) What direction is the bomb moving just before it hits thetarget?
2° below thehorizontal. this not the correct answer and I have no ideahow to get it.

Please someone help me understand how to get the correctanswer.

Explanation / Answer

       Given that the initialhorizontal velocity is U = 38.6 m/s        The initial height is H =100 m     -----------------------------------------------------------------    The horizontal distance covered by the bomb is R= horizontal velocity*time of flight                                                                               = U*2h/g                                                                               = 38.6m/s*(2*100m / 9.8m/s2)                                                                                =174.4 m    The horizontal velocity after time t is Ux = U =38.6 m/s The verticle velocity after time t is Uy = 0+ gt = g*2h/g      (since V = U +at)                                                            =2gh                                                              =44.27 m/s     The angle is =tan-1(Uy/Ux)                             =48.910   

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