Question

A bullet is fired from a rifle emerging from the muzzle at 340m/s. it strikes a sandbag some distance away, having lost 10 percent of its velocity due to air resistance. If it penetrates the sand bag to a depth of 12centimeters how long did it take for the bullet to come to rest in the sand bag

Answer 1

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

Answer 2

Answer:

8×10^-4

Explanation:

10% of 340 =306m/s

u(initial velocity) =306m/s

u(initial velocity) =306m/ss=12 cm=0.12m

u(initial velocity) =306m/ss=12 cm=0.12mv(final velocity) =0m/s (Stops to rest)

u(initial velocity) =306m/ss=12 cm=0.12mv(final velocity) =0m/s (Stops to rest) Acceleration=?

u(initial velocity) =306m/ss=12 cm=0.12mv(final velocity) =0m/s (Stops to rest) Acceleration=?Time=?

u(initial velocity) =306m/ss=12 cm=0.12mv(final velocity) =0m/s (Stops to rest) Acceleration=?Time=?According to 3rd equation,

u(initial velocity) =306m/ss=12 cm=0.12mv(final velocity) =0m/s (Stops to rest) Acceleration=?Time=?According to 3rd equation, v^2=u^2+2as

u(initial velocity) =306m/ss=12 cm=0.12mv(final velocity) =0m/s (Stops to rest) Acceleration=?Time=?According to 3rd equation, v^2=u^2+2as0=306^2 + 2(0.12) a

u(initial velocity) =306m/ss=12 cm=0.12mv(final velocity) =0m/s (Stops to rest) Acceleration=?Time=?According to 3rd equation, v^2=u^2+2as0=306^2 + 2(0.12) a0=93636 +0.24a

+0.24a-93636=0.24a

=0.24a-93636/0.24=a

/0.24=a-390150m/s^2=a

=aAccording to 1st equation,

=aAccording to 1st equation, v=u+at

=aAccording to 1st equation, v=u+at0= 306 + (-390150)t

)t-306/(-390150) =t

) =tt= 0.000784313725

t = 0.0008 (approximately)

t= 8×10^-4