a bullet is fired horizontally froma height of 180m. The bullet undergoes vertical and horizontal displacement of 500m and 45m respectively in nth second.What will be maximum horizontal distance travel by the bullet.(take g=9.8)
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Explanation:
500 = 4.9(2n-1) , n = (500/4.9 +1) ÷2 = 51.52
45 = u ( n - n +1 ) = u
180 = 4.9 t² , t = 6.06
max. horizontal distance = 45x6.06 = 272.7 m
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