A bullet going with speed 350m/S enters a concrete wall and penetrates a distance of 50 cm before coming to rest.find the deceleration
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u = 350 m/s
v = 0
s = 50 cm = 0.5 metres
By using the third equation of uniformly accelerated motion,
we have,
v² = u²+2as
0² = 350² + 2×a×0.5
-122500 = a
a = -122500 m/s²
deceleration of bullet will be 122500 m/s².
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v = 0
s = 50 cm = 0.5 metres
By using the third equation of uniformly accelerated motion,
we have,
v² = u²+2as
0² = 350² + 2×a×0.5
-122500 = a
a = -122500 m/s²
deceleration of bullet will be 122500 m/s².
_PLEASE MARK IT AS BRAINLIEST ANSWER AND DROP A ♥_
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