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Answer :
Final answer:
The coordinates of the shell where it explodes are x = 6909.4 m, y = 6376.6 m. Just before the shell hits the mountainside, the x and y components of its velocity are 164 m/s and -50.78 m/s, respectively.
Explanation:
This problem involves using the principles of projectile motion, in which an object is launched into the air and subject to only the acceleration due to gravity. We can decompose the projectile's motion into horizontal and vertical components.
For the x-coordinate (a), remember that the horizontal motion is constant, so we use the formula x = vt, where v is the horizontal component of the initial velocity, and t is time. Since the shell is fired at an angle, we use trigonometry to find this horizontal component, i.e., vx = v0cos(θ) = 328 m/s * cos(60.0 degrees) = 164 m/s. Multiplying this by time (t = 42.1 s), we get x = 6909.4 m.
For the y-coordinate (b), it's the vertical displacement of the projectile. The vertical motion is influenced by gravity, so we use the formula y = v0sin(θ)t - 0.5gt2. Here, v0sin(θ) is the vertical component of the initial velocity, g is the acceleration due to gravity (9.8 m/s2) and t is time. By plugging in the values, we get y = 328 m/s * sin(60 degrees) * 42.1 s - 0.5 * 9.8 m/s2 * (42.1 s)2 = 6376.6 m.
For the x component of velocity (c), it remains constant, so it is the same as the initial horizontal velocity: 164 m/s.
For the y component of velocity (d), we use the formula v = v0sin(θ) - gt to account for the acceleration due to gravity. That equates to 328 m/s * sin(60 degrees) - 9.8 m/s2 * 42.1 s = -50.78 m/s.
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Rewritten by : Brahmana