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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`
4 months ago

Saurabh Koranglekar
3163 Points
```							Dear studentSpeed after decrease of 10 % = 306 m/sFinal speed = 0a = (306*306)/(0.12)Time t = u/a= 0.12/306 = 0.39 mili secondsRegards
```
4 months ago
Khimraj
3008 Points
```							U1=340m/sS from striking bag= 12 cm=0.12 mV=0 So, when striking, u= 340-(340×1/100)= 306m/sa= (0-306×306)÷(2×0.12)=-51×765m/s×sV=U+ator, 0=306+ (-51×765)×tor, -306÷(-51×765=tor, t= 0.0078431372 second or, 0.008 second
```
4 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions