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Grade: 9
        
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
4 months ago

Answers : (2)

Saurabh Koranglekar
askIITians Faculty
3163 Points
							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
4 months ago
Khimraj
3008 Points
							
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
4 months ago
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