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 = 0.44 m, and x = 6.8 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.

Question

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 = 0.44 m, and x = 6.8 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.

Answer 1

I have a very good question too. With no drawing, how in the world do we even know what you are talking about with distances D and H? We need to be mind readers? Some teachers may be able to do that but I'm not one of them.

Answer 2

To determine the distances D and H, we can analyze the motion of the bullet and use the information provided in the drawing.

First, let's break down the motion into horizontal and vertical components. Since the bullet is fired parallel to the ground, there is no initial vertical velocity. Therefore, the vertical motion can be treated as free fall.

Using the information provided, we know that the bullet hits the wall at a height y = 0.44 m above the ground. This means that the time it takes for the bullet to reach that height is the same as the time it takes to travel horizontally.

We can determine the time it takes for the bullet to reach the window using the horizontal distance x = 6.8 m and the initial horizontal velocity of the bullet, which is equivalent to the speed of 340 m/s.

Time = Distance / Speed
t = x / v

Plugging in the values:
t = 6.8 m / 340 m/s = 0.02 s

Since the vertical motion is free fall, we can use the equation of motion to calculate the vertical distance traveled by the bullet:

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

Where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Plugging in the values:
0.44 m = (1/2) * 9.8 m/s^2 * (0.02 s)^2

Simplifying:
0.44 m = 0.0049 m/s^2 * 0.0004 s^2

Now, we can determine the vertical component of the initial velocity (V0y) using the equation:

V0y = g * t

Plugging in the values:
V0y = 9.8 m/s^2 * 0.02 s = 0.196 m/s

Now that we have determined the initial vertical velocity, we can find the total initial velocity (V0) using the Pythagorean theorem:

V0 = sqrt(V0x^2 + V0y^2)

Since the bullet is fired parallel to the ground, there is no change in horizontal velocity, so V0x remains constant and equal to 340 m/s.

Plugging in the values:
V0 = sqrt((340 m/s)^2 + (0.196 m/s)^2) = 340.0004 m/s

Now we have enough information to calculate the distances D and H.

D is the horizontal distance between the point of firing and the window. It is determined by the initial horizontal velocity V0x and the time t:

D = V0x * t = 340 m/s * 0.02 s = 6.8 m

H is the vertical distance between the point of firing and the window. It is determined by the initial vertical velocity V0y and the time t:

H = V0y * t = 0.196 m/s * 0.02 s = 0.00392 m

Therefore, the distances D and H are approximately 6.8 m and 0.00392 m, respectively.