NEET Class neet Answered
A bomber moving horizontally with 500m/s drops a bomb which strikes ground in 10 s. The angle of strike with horizontal is
Asked by Vinasadariya | 03 Sep, 2019, 07:50: PM
Expert Answer
Vertical velocity component acquired by bomber in 10 s = g×t = 9.8×10 = 98 m/s
Horizontal component of velocity will remain same as there is no acceleration in horizontal direction
Horizontal component of velocity = 500 m/s
hence angle of strike with horizontal =tan-1 (Vv/VH )=tan-1 (98/500) = 11°
Answered by Utkarsh Lokhande | 03 Sep, 2019, 10:08: PM
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