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A gun is fired with angle of elevation 30. What is the muzzle speed if the maximum height of the shell is 440 m?
Wiki User
â 12y ago
Y=(Vo^2 x sin^2(theta))/2g
Solve for Vo
Wiki User
â 12y ago
The maximum height of the shell occurs when the vertical component of velocity is zero. Using the kinematic equation v_f^2 = v_i^2 + 2ad where a is -9.8 m/s^2, solve for the initial velocity v_i, which is the muzzle speed. The equation becomes 0 = v_i^2 * sin^2(30) - 2 * g * h_max, where h_max = 440 m, giving a muzzle speed of approximately 227 m/s.
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How does angle of projection affect the maximum height?
The angle of projection affects the maximum height by determining the vertical and horizontal components of the initial velocity. At 90 degrees (vertical), all the initial velocity is vertical which results in maximum height. As the angle decreases from 90 degrees, the vertical component decreases, leading to a lower maximum height.
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