From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. The bullet puts a hole in a window of another building and hits the wall that faces the window. (y = 0.42 m, and x = 7.2 m.)Determine the distances D and H, which locate the point where the gun was fired. Assume that the bullet does not slow down as it passes through the window
To determine the distances D and H, we can use the horizontal and vertical components of the bullet's motion.
Let's analyze the horizontal motion first. We know the horizontal distance the bullet traveled, which is x = 7.2 m. The horizontal component of the bullet's velocity remains constant at 340 m/s throughout its motion because there are no horizontal forces acting on it. Therefore, we can use the formula:
x = Vx * t
where Vx is the horizontal component of the bullet's velocity and t is the time it took for the bullet to travel the distance x.
Rearranging the formula, we can solve for time:
t = x / Vx
t = 7.2 m / 340 m/s
t ≈ 0.0212 s
Now, let's consider the vertical motion. We know the vertical coordinate, y = 0.42 m. The bullet experiences acceleration due to gravity (g = 9.8 m/s²) in the vertical direction. Using the equation of motion:
y = Vy₀ * t + (1/2) * g * t²
where Vy₀ is the initial vertical velocity (which is equal to 0 since the bullet was fired parallel to the ground).
Substituting the values:
0.42 m = 0 * t + (1/2) * (9.8 m/s²) * (0.0212 s)²
Simplifying the equation and solving for t²:
t² ≈ (2 * 0.42 m) / (9.8 m/s²)
t² ≈ 0.086 m / (9.8 m/s²)
t² ≈ 0.0088 s²
Now, knowing the time it took for the bullet to reach the window, we can determine the distances D and H.
D = Vx * t
D ≈ 340 m/s * 0.0212 s
D ≈ 7.208 m
Therefore, the distance D, which locates the point where the gun was fired, is approximately 7.208 meters.
To calculate the height H, we can use the equation of motion again:
H = Vy * t + (1/2) * g * t²
Since Vy is the vertical component of the bullet's velocity and the bullet was fired parallel to the ground, Vy remains constant at 0 m/s.
H = 0 * t + (1/2) * (9.8 m/s²) * (0.0212 s)²
H = 0.5 * (9.8 m/s²) * (0.00044944 s²)
H ≈ 0.0219 m
Therefore, the height H, which locates the point where the gun was fired, is approximately 0.0219 meters.
To determine the distances D and H, we can use the horizontal and vertical components of the bullet's motion.
Given:
Initial speed of the bullet, v = 340 m/s
Vertical displacement, y = 0.42 m
Horizontal displacement, x = 7.2 m
Step 1: Determine the time of flight
The time of flight can be found using the vertical motion of the bullet. We can use the equation:
y = v0y * t + (1/2) * g * t^2
Where:
v0y is the initial vertical velocity (0 since the bullet was fired parallel to the ground),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time of flight.
Plugging in the given values:
0.42 = 0 * t + (1/2) * 9.8 * t^2
0.42 = 4.9 * t^2
Solving for t:
t^2 = 0.42 / 4.9
t^2 = 0.0857
t ≈ √0.0857
t ≈ 0.293 s (rounded to 3 decimal places)
Step 2: Determine the horizontal distance D
The horizontal distance D can be found using the formula:
D = v0x * t
Where:
v0x is the initial horizontal velocity (equal to the speed of the bullet, 340 m/s),
t is the time of flight.
Plugging in the values:
D = 340 * 0.293
D ≈ 99.32 m (rounded to 2 decimal places)
Step 3: Determine the height H
The height H can be found using the formula:
H = v0y * t - (1/2) * g * t^2
Plugging in the values:
H = 0 * 0.293 - (1/2) * 9.8 * 0.293^2
H ≈ -0.432 m (rounded to 3 decimal places)
Since the height cannot be negative, we can consider the bullet to be fired below the initial height of the window.
Therefore, the distances D and H that locate the point where the gun was fired are approximately D = 99.32 m and H = -0.432 m, respectively (rounded to the appropriate decimal places).