1.It is possible to shoot an arrow at a speed as high as 103 m/s. (a) If frictio
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Question
1.It is possible to shoot an arrow at a speed as high as 103 m/s.
(a) If friction is neglected, how high would an arrow launched at this speed rise if shot straight up?
(a) Determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel. (Use tas necessary and round all numerical coefficients to exactly 3 significant figures.)
(b) Determine the length of time the bullet is accelerated.
s
(c) Find the speed at which the bullet leaves the barrel.
m/s
(d) What is the length of the barrel?
m
Explanation / Answer
Q1.
part a:
initial speed=1000 m/s
as there is no friction involved, total mechanical energy will be conserved.
then initial potential energy+initial kinetic energy=final potential energy+final kinetic energy
final kinetic energy is 0 as at maximum height, speed=0 m/s
gravitational potential energy is given as -G*M*m/r
where G=gravitational constant=6.674*10^(-11)
M=mass of earth=5.972*10^24 kg
m=mass of the arrow
r=distance from center of earth
R=radius of earth=6.371*10^6 m
let maximum height be h.
then writing energy balance equation:
(-G*M*m/R)+0.5*m*v^2=(-G*M*m/(R+h))
==>0.5*v^2=G*M*((1/R)-(1/(R+h))
==>0.5*10^6=6.674*10^(-11)*5.972*10^24*((1/6371000)-(1/(R+h))
==>h=5.1329*10^4 m
so maximum height achieved by such arrow is 5.1329*10^4 m
Q2. v=-4.75*10^7*t^2+2.75*10^5*t
acceleration=a=dv/dt=-9.5*10^7*t+2.75*10^5
distance=integration of v*dt
x=-1.583*10^7*t^3+1.375*10^5*t^2
part b:
just as it reaches the the end of the barrel, acceleration becomes zero.
==>-9.5*10^7*t+2.75*10^5 =0
==>t=2.75*10^5/(9.5*10^7)
=2.895*10^(-3) s
part c:
speed at t=2.895*10^(-3) seconds is given by
v=-4.75*10^7*t^2+2.75*10^5*t
=398 m/s
part d:
length of the barrel=value of x at t=2.895*10^(-3) seconds
=-1.583*10^7*t^3+1.375*10^5*t^2
=0.7682 m
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