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Chapter 7: Problem 6

A shell is fired from a gun with a muzzle velocity of \(466 \mathrm{~m} / \mathrm{s}\), at an angle of \(57.4^{\circ}\) with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming level terrain?

Short Answer

Expert verified
The other fragment of the shell lands at a distance equal to twice the horizontal distance before the explosion. To find the specific value, finish the calculations as explained in the steps.

Step by step solution

01

Determine the vertical and horizontal components of the initial velocity

We can resolve the initial velocity of 466 m/s into two components: horizontal and vertical. The horizontal component (v_{x0}) can be found using the equation \(v_{x0} = v_0 \cdot cos(\Theta)\), and the vertical component (v_{y0}) can be found using the equation \(v_{y0} = v_0 \cdot sin(\Theta)\) where \(\Theta\) is the angle of projection, 57.4 degrees.
02

Calculate the time to reach the greatest height

At the highest point of the trajectory, the vertical speed is zero. Using the kinematic equation, \(v_{y} = v_{y0} - g \cdot t\), where g is the acceleration due to gravity (9.81 m/s²), time (t) to reach the highest point can be calculated as \(t = v_{y0} / g\).
03

Calculate the horizontal distance before the explosion

The horizontal distance (d) before the explosion can be calculated using the simple formula \(d = v_{x0} \cdot t\), as in the x-direction there is no acceleration.
04

Calculate the horizontal distance after the explosion

After the explosion, due to conservation of momentum, the second fragment continues with the same horizontal speed. Therefore, we calculate the total time it takes for the fragment to hit the ground, and then calculate the total horizontal distance it covers using the formula \(d = v_{x0} \cdot T\), where T is the total time (time to go up and come down).

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