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

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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 = 0.60 m, and x = 7.1 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.

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Ostrovityanka [3.2K] · Answer 1 · 2026-04-11T03:05:23+03:00

The point of launch is determined to be a distance of D = 962.2 m and H = 39.2 m. Explanation: It is essential to visualize the situation with a diagram to ease understanding. This is a projectile launching scenario with specified parameters after passing through the window and impact on the wall; we compute the contact velocities. For the X-axis: x = Vox t, thus t = x / vox leads to t = 7.1 / 340, which gives t = 2.09 x 10⁻² s. During this timeframe, the height drop is calculated using Y = Voy t - ½ g t², allowing us to derive Voy. Applying Voy = (Y + ½ g t²) / t gets us Voy = [0.6 + ½ 9.8 (2.09 x 10⁻²)²] / 2.09 x 10⁻² = 27.7 m/s. Next, we establish the distances D and height H from the launch position, determining the time taken to move toward the window, given Vy2 = 27.7 m/s. Time t2 = Vy / g results in t2 = 27.7 / 9.8 leading to t2 = 2.83 s. This is the same duration that passes for both horizontal and vertical travel: X = Vox t2, thus D = 340 * 2.83 gives D = 962.2 m. For height, applying Y = Voy² - ½ g t2² allows the conclusion that H = - ½ 9.8 * (2.83)² fixes H = 39.2 m.