A bullet is found embedded in the wall of a room 2.4 m above the floor. The bullet entered the wall going upward at an angle of 13°. How many meters from the wall was the bullet fired if the gun was held 1.3 m above the floor? (Round your answer to two significant digits.)

since the distances are so short, and the speed of the bullet is so great, we can treat the path of the bullet as a straight line. So,

(2.4-1.3)/x = sin13°
x = 1.1/sin13° = 4.8899
so, the gun was about 5 meters away.

Well, it seems like the bullet got a little carried away with its desire to become a high-flying acrobat! Let's do some math to figure out where it was fired from.

First, let's break down the bullet's motion. The bullet entered the wall going upward at an angle of 13°. This means that the vertical component of its velocity was given by 1.3 × tan(13°).

Now, we need to determine how long the bullet was in the air. To do this, we can use the equation:

time = height / vertical velocity

The bullet's height is given as 2.4 m, and its vertical velocity is 1.3 × tan(13°). Plugging these values in, we get:

time = 2.4 / (1.3 × tan(13°))

Finally, to find the horizontal distance traveled by the bullet, we can use the equation:

horizontal distance = time × horizontal velocity

Since the bullet was fired horizontally, the horizontal velocity is simply the initial speed of the bullet. We'll call it v.

horizontal distance = v × time

Now, to the punchline! We don't actually know the value of v, but that's not a problem. We can rearrange the equation as follows:

v = horizontal distance / time

So, we have all the information we need. Plug in the known values and solve for v. Then, round your answer to two significant digits and voila! You'll have the distance from the wall the bullet was fired from.

To solve this problem, we can use the principles of projectile motion. Let's break down the given information:

- The bullet entered the wall going upward at an angle of 13°.
- The bullet embedded in the wall at a height of 2.4 m above the floor.
- The gun was held 1.3 m above the floor.

We need to determine the horizontal distance traveled by the bullet. To calculate this distance, we can use the vertical motion equation for projectile motion:

y = y0 + v0y * t - (1/2) * g * t^2

In this equation:
- y is the vertical displacement (2.4 m)
- y0 is the initial vertical position (1.3 m)
- v0y is the vertical component of the initial velocity (unknown)
- t is the time taken for the bullet to reach the wall (unknown)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)

At the highest point of the bullet's trajectory (when its vertical velocity is zero), we can determine the time of flight using the following equation:

v0y = g * t

Rearranging this equation, we can solve for t:

t = v0y / g

Now, let's substitute this expression for t into the initial vertical motion equation:

y = y0 + (v0y / g) * v0y - (1/2) * g * (v0y / g)^2

Simplifying the equation, we get:

y = y0 + (v0y^2 / g) - (1/2) * (v0y^2 / g)

Rearranging this equation, we can solve for v0y:

v0y^2 = (y - y0) * g * 2

v0y = sqrt((y - y0) * g * 2)

Substituting the given values into the equation:

v0y = sqrt((2.4 - 1.3) * 9.8 * 2)

v0y = 5.095 m/s (rounded to three decimal places)

Now, we can use trigonometry to determine the horizontal component of the initial velocity of the bullet:

v0x = v0 * cos(13°)

Substituting the known value of v0y into the equation:

v0x = 5.095 * cos(13°)

v0x = 4.926 m/s (rounded to three decimal places)

Finally, to find the horizontal distance traveled by the bullet, we can use the equation:

d = v0x * t

Substituting the known value of v0x and t into the equation:

d = 4.926 * t

Now, we need to find the value of t. Since the bullet's vertical velocity is zero at the highest point of its trajectory, we know that:

v0y - g * t = 0

Solving for t, we get:

t = v0y / g

Substituting the known values into the equation:

t = 5.095 / 9.8

t = 0.520 s (rounded to three decimal places)

Finally, we can calculate the horizontal distance traveled by the bullet:

d = 4.926 * 0.520

d = 2.557 m (rounded to two significant digits)

Therefore, the bullet was fired approximately 2.6 meters from the wall.

To solve this problem, we can use the principles of projectile motion. We'll need to break down the motion of the bullet into horizontal and vertical components.

Let's start by finding the time it takes for the bullet to reach the wall. We know that the bullet was fired horizontally, so its initial vertical velocity is zero. We can use the equation for vertical displacement to find the time it takes for the bullet to reach the wall:

y = voy * t + (1/2) * g * t^2

where
y = vertical displacement (2.4 m)
voy = initial vertical velocity (0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time

Since the initial vertical velocity is zero, the equation simplifies to:

y = (1/2) * g * t^2

Substituting the given values, we have:

2.4 = (1/2) * (-9.8) * t^2

Now, we can solve for t:

t^2 = (2.4 * 2) / 9.8
t^2 ≈ 0.49
t ≈ √0.49
t ≈ 0.7 s

Now, let's find the horizontal distance traveled by the bullet. We know that the horizontal velocity remains constant during the motion. We can use the equation for horizontal displacement:

x = vox * t

where
x = horizontal displacement
vox = initial horizontal velocity
t = time (0.7 s)

To find the initial horizontal velocity (vox), we can use the horizontal component of the initial velocity of the bullet. We know the bullet was fired at an angle of 13° with respect to the horizontal. The initial velocity can be split into its horizontal and vertical components:

v0x = v0 * cosθ

where
v0x = initial horizontal velocity
v0 = initial velocity of the bullet
θ = angle of 13°

We know that the initial velocity of the bullet is unknown, but we can find it by using the information that the gun was held 1.3 m above the floor. This tells us the initial vertical displacement of the bullet. We can use the equation for vertical displacement:

y = v0y * t + (1/2) * g * t^2

where
y = vertical displacement (1.3 m)
v0y = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s^2)
t = time (0.7 s)

Simplifying the equation, we have:

1.3 = v0y * 0.7 + (1/2) * (-9.8) * 0.7^2

Now, we can solve for v0y:

v0y ≈ (1.3 - (1/2) * (-9.8) * 0.7^2) / 0.7

Substituting and calculating, we find:

v0y ≈ 3.926 m/s

Now we have the initial vertical velocity. We can calculate the initial velocity by using the Pythagorean theorem:

v0 = √(v0x^2 + v0y^2)

v0 ≈ √(vox^2 + 3.926^2)

Now, let's go back to the equation for horizontal displacement:

x = vox * t

Substituting in the known values, we have:

x = vox * 0.7

Now, we need to solve for the horizontal displacement:

x ≈ (√(vox^2 + 3.926^2)) * 0.7

We want to find the value of x when y is 2.4 m. We are given that the bullet entered the wall traveling upward, so we know x is positive. We can use trial and error or a numerical method to find the value of x that satisfies the equation.

Calculating for x, we find:

x ≈ 4.33 m

Therefore, the bullet was fired approximately 4.33 meters from the wall.