A bullet is fired from a rifle emerging from the muzzle At 340 metres per second it strikes a sandbag some distance away having lost 10% of its velocity due to air resistance if it penetrates the sand bag to a depth of 12cm how long does it take for the bullet to come at the rest in the sand bag

Question

Saurabh Koranglekar · Answer

Dear student Speed after decrease of 10 % = 306 m/s Final speed = 0 a = (306*306)/(0.12) Time t = u/a= 0.12/306 = 0.39 mili seconds Regards

Khimraj · Answer

U1=340m/s
S from striking bag= 12 cm=0.12 m
V=0
So, when striking, u= 340-(340×1/100)= 306m/s
a= (0-306×306)÷(2×0.12)=-51×765m/s×s
V=U+at
or, 0=306+ (-51×765)×t
or, -306÷(-51×765=t
or, t= 0.0078431372 second
or, 0.008 second

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A bullet is fired from a rifle emerging from the muzzle At 340 metres per second it strikes a sandbag some distance away having lost 10% of its velocity due to air resistance if it penetrates the sand bag to a depth of 12cm how long does it take for the bullet to come at the rest in the sand bag

A bullet is fired from a rifle emerging from the muzzle At 340 metres per second it strikes a sandbag some distance away having lost 10% of its velocity due to air resistance if it penetrates the sand bag to a depth of 12cm how long does it take for the bullet to come at the rest in the sand bag

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A bullet is fired from a rifle emerging from the muzzle At 340 metres per second it strikes a sandbag some distance away having lost 10% of its velocity due to air resistance if it penetrates the sand bag to a depth of 12cm how long does it take for the bullet to come at the rest in the sand bag

A bullet is fired from a rifle emerging from the muzzle At 340 metres per second it strikes a sandbag some distance away having lost 10% of its velocity due to air resistance if it penetrates the sand bag to a depth of 12cm how long does it take for the bullet to come at the rest in the sand bag

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