23. A rifle is aimed horizontally toward the center of a target 71.4 m away. If the bullet strikes 10 cm below the center, what was the velocity of the bullet?

First calculate how long (T) it takes the bullet to fall 10 cm, with zero initial vertical velocity component.

(1/2) g T^2 = 0.1 meter
T = 0.143 s

Then solve V*T = 71.4 m, to get V

To find the velocity of the bullet, we can use the principles of projectile motion.

The horizontal velocity of the bullet remains constant throughout its flight. This means that the horizontal component of the bullet's velocity is the same as the initial velocity of the bullet, given that it was aimed horizontally.

Let's assume that the horizontal velocity of the bullet is v_x.

The vertical motion of the bullet is influenced by acceleration due to gravity. The vertical position of the bullet can be described using the equation:

y = v_y * t - (1/2) * g * t^2,

where y is the vertical distance traveled by the bullet, v_y is the vertical component of the bullet's velocity, g is the acceleration due to gravity, and t is the time of flight.

Since the bullet is aimed horizontally, the initial vertical velocity (v_y) of the bullet is zero.

The time of flight (t) can be determined using the equation:

y = v_y * t - (1/2) * g * t^2.

In this case, the vertical distance traveled is -0.1 m (since the bullet strikes 10 cm below the center of the target), v_y is zero, and g is approximately 9.8 m/s^2.

Therefore:

-0.1 = 0 - (1/2) * 9.8 * t^2.

Simplifying the equation, we get:

0.1 = 4.9 * t^2.

Now, solve for t:

t^2 = 0.1 / 4.9.

t = sqrt(0.1 / 4.9).

t ≈ 0.0451 seconds.

Now, we can use the time of flight to find the horizontal velocity of the bullet (v_x) using the equation:

d = v_x * t.

In this case, the horizontal distance traveled (d) is 71.4 m.

Therefore:

71.4 = v_x * 0.0451.

Simplifying the equation, we get:

v_x = 71.4 / 0.0451.

v_x ≈ 1585 m/s.

Hence, the velocity of the bullet was approximately 1585 m/s.