What was the bullet's speed as it left the barrel?

Question

A rifle is aimed horizontally at a target 55.0 m away. The bullet hits the target 2.60 cm below the aim point. What was the bullet's flight time? What was the bullet's speed as it left the barrel?

Answer 1

v=bulltet's speed at leaving the barrel -- along the horizontal

v_x = x -component of v = v
v_y = y -component of v = 0

Let u = bullet's velocity when it hits the target
Since the distance is not much, the bullet does not loose velocity along the horizontal direction due to air resistance.

Hence , u_x = x-component of bullet's final velocity = v_x =v

(u_y)^2 = (v_y)^2 +2*f*s
= 0 + 2 * g * 2.6 cm
= 2 * 9.8 m/sec^2 * 2.6*10^-2m
= 5.096e-1 (m/sec)^2
= 0.5096 (m/sec)^2

Hence u_y = sqrt(5.096e-1) m/s
=0.713863 m/sec

Hence the speed of the bullet as it hits the target
u = sqrt [ (u_y)^2 + (u_x)^2 ]
since u_y and u_x are normal to each other and we use the law of addition of vector and find its magnitude.

or, u = sqrt [ v^2 + 0.5096 ] m/sec

Again, we have,
u_y = v_y + g * t

[ all y-coordinates, velocities and acceleration positive downwards]
u_y = 0 + 9.8 *t
Hence, time of flight
t = 0.713863 /9.8 sec
= 0.072843 sec

Now since u_x = v_x (as there is no or negligible force in the x-direction), we have

d= 55.0 m = v_x * t
v_x = v = (55.0/0.072843 ) m/sec
= 755.047 m/sec
Hence we get

u = sqrt[ (u_x)^2 + (u_y)^2]
= 755.047458 m/sec

Answers:
What was the bullet's flight time? - 0.072843 sec

What was the bullet's speed as it left the barrel?
-- 755.047 m/sec

Cheers.

Answer 2

Ignoring all different elements and assuming that the rifle is aimed properly, and so on., the drop decrease than aimpoint tells you the bullt fell 2 cm in the course of its flight. because the vertical habit of the bullet is in undemanding words laid low with gravity, you may make certain how lengthy it took to fall both cm. The time calculated for the fall is a similar time it took to commute 50 m, so that you'll be able to calculate the horizontal speed of the bullet.

Answer 3

Time to drop 2.6 cm:
t = (2x/g)^0.5 = (2 * 2.6/980)^0.5 = 0.073 sec

Muzzle velocity (which doesn't need to be distinguished from any horizontal velocity since it won't change significantly even with air resistance effects accounted for):
v = 5500/0.073 = 75300 cm/sec = 753 m/sec (to 3 sig figs)