A rifle is aimed horizontally at a target 50 m away. The bullet hits the target 2.4 cm below the aim point

What was the bullet's speed as it left the barrel?
Express your answer to two significant figures and include the appropriate units

To calculate the bullet's speed as it left the barrel, we can use the principles of projectile motion.

First, we need to find the time it takes for the bullet to travel horizontally to the target. We know that the horizontal distance is 50m, and since the bullet is traveling horizontally, we can ignore the effect of gravity in this direction. The formula for time is:

time = distance / velocity

Since the velocity in the horizontal direction is constant, we can use the given distance of 50m. Therefore:

time = 50m / velocity

Next, we can find the vertical distance that the bullet drops below the aim point. The bullet hits the target 2.4cm below the aim point. We know that the bullet is subject to the force of gravity in the vertical direction, so we can use the formula for vertical displacement:

displacement = (1/2) * acceleration * time^2

Since the bullet is accelerating due to gravity with a value of 9.8m/s^2, and the vertical displacement is -2.4cm (-0.024m), we can rewrite the equation as:

-0.024m = (1/2) * 9.8m/s^2 * time^2

Now, we have two equations with two variables (time and velocity). We can solve these equations simultaneously to find the bullet's velocity as it left the barrel.

Let's start by rearranging the second equation to solve for time:

time^2 = (-0.024m) / [(1/2) * 9.8m/s^2]

time^2 = -0.024m / 4.9m/s^2

time^2 = -0.0049s^2

time ≈ 0.07s

Now we can substitute this value of time into the first equation:

0.07s = 50m / velocity

Rearranging the equation to solve for velocity:

velocity = 50m / 0.07s

velocity ≈ 714m/s

Therefore, the bullet's speed as it left the barrel is approximately 714 m/s.

h = 0.5g*t^2 = 0.024 m.

4.9*t^2 = 0.024.
t = ?

Dx = Xo^t = 50 m.
Xo*t = 50.
Xo = ?.

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