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. As the drawing shows, the bullet puts a hole in a window of another building and hits the wall that faces the window. (y = 0.56 m, and x = 7.0 m.) Using the data in the drawing, 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.
The time that the bullet spends in the building
t=x/v(0x) =7/340 =0.02 s.
The vertical displacement of the bullet in the building y= 0.56 m.
The vertical component of the velocity of the bullet as it passes through the window is
v(y) = (y- gt²/2)/t = y/t - gt/2=
=0.5/0.02 -9.8•0.02/2 =24.9 m/s
y=v(y)²/2g= 24.9²/2•9.8 = 31.6 m
H=31.6+0.56 =32.16 m
The time for the bullet to reach the window is
t₀ =2y/v(y) = 2•31.6/24.9 = 2.54 s
D=v(0x)t₀ = 340•2.54= 863.6 m
This is not the answer
To solve this problem, we can use the equations of motion for projectile motion. We know that the bullet was fired parallel to the ground, so it will only experience horizontal motion.
Let's break down the given information:
- The horizontal distance from the window to the point where the bullet was fired is x = 7.0 m.
- The vertical distance from the window to the point where the bullet was fired is y = 0.56 m.
- The initial velocity of the bullet is 340 m/s.
We can use the equation for horizontal motion to determine the time it took for the bullet to travel a horizontal distance of x:
x = v * t
where x is the horizontal distance, v is the initial velocity, and t is the time. Plugging in the known values:
7.0 m = 340 m/s * t
Solving for t:
t = 7.0 m / 340 m/s
t ≈ 0.0206 s
Now that we know the time, we can use the equation for vertical motion to determine the vertical distance the bullet traveled:
y = (1/2) * g * t^2
where y is the vertical distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Plugging in the known values:
0.56 m = (1/2) * 9.8 m/s^2 * (0.0206 s)^2
Solving for y:
y ≈ 1/2 * 9.8 m/s^2 * (0.0206 s)^2
y ≈ 0.00528 m
Now we can determine the distances D and H.
D is the horizontal distance between the point where the bullet was fired and the window. Since the bullet travels horizontally at a constant velocity, D is equal to the horizontal distance traveled in time t:
D = v * t
Plugging in the known values:
D = 340 m/s * 0.0206 s
D ≈ 7.004 m
H is the vertical distance between the point where the bullet was fired and the window. Since the bullet experiences vertical freefall motion with an initial velocity of 0 m/s, H can be calculated using:
H = y
Plugging in the known values:
H ≈ 0.00528 m
Therefore, the distances D and H are approximately 7.004 m and 0.00528 m, respectively.
To determine the distances D and H, which locate the point where the gun was fired, we can use the kinematic equations of motion.
First, let's examine the vertical motion of the bullet. Since the bullet leaves the gun with a speed of 340 m/s parallel to the ground, there is no initial vertical velocity. We can use the equation:
y = vit + (1/2)gt^2
Where:
y = vertical displacement (0.56 m)
vi = initial vertical velocity (0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time of flight
Since the bullet takes the same time to travel horizontally as it does vertically, we can use the time of flight for both equations.
Substituting the given values into the equation, we can solve for t:
0.56 = 0*t + (1/2)(-9.8)t^2
Simplifying the equation:
0.56 = (-4.9)t^2
t^2 = 0.56 / -4.9
t^2 = -0.1143
Since time cannot be negative, we discard the negative result. Therefore, t^2 = 0.1143.
Taking the square root of both sides:
t ≈ 0.338 second
Now, let's examine the horizontal motion of the bullet. The horizontal distance traveled can be calculated using the equation:
x = vix * t
Where:
x = horizontal distance (7.0 m)
vix = initial horizontal velocity (340 m/s)
t = time of flight (0.338 s)
Substituting the given values into the equation:
7.0 = 340 * 0.338
Simplifying the equation:
7.0 = 114.92
Therefore, the horizontal distance D, which locates the point where the gun was fired, is approximately equal to 114.92 meters.
To determine the vertical distance H, we can use the equation:
H = viy * t
Since there is no initial vertical velocity, viy = 0.
Hence, the vertical distance H is equal to 0 meters.
Therefore, the distances D and H, which locate the point where the gun was fired, are approximately 114.92 meters and 0 meters, respectively.