Question

A bullet is fired from a rifle, emerging from he muzzle at 340 metres per second. It strikes a sandbag some distance away, having lost
10 percent of its velocity due to air resistance. If it penetrates the sandbag to a depth of
12.0 centimeters, how long did it take for the bullet to come to rest in the sandbag?​

Answer 1

Answer:

U1=340 m/s

S from stricking bag =12cm =0.12 m

v=0

so,when stricking, u =340 -(340×1/100) = 306 m/s

a=(0.306×306)÷(2×0.12) =.51 ×765m/s

V=U+at

or,0 =306+(-51×765)×t

or, +306 ÷ ( -51 × 765 )=t

or, t =0.00784313725 second

or, 0.008 second