HardPhysicsfluid
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-- 2. A body is projected with certain velocity from ground. At a height of 0.4 m from the ground, the velocity of a body is v = (6î + 2ĵ) m/s then the angle of projection from horizontal is (g = 10 m/s²)
A
30°
B
60°
C
45°
D
tan⁻¹(1/3)
Explanation
v_x = 6 (constant). u_x = 6. At y=0.4, v_y = 2. Using v_y² = u_y² - 2gy: 2² = u_y² - 2(10)(0.4) => 4 = u_y² - 8 => u_y² = 12 => u_y = √12 = 2√3 (approx). Note: If ĵ component was different or height was different, results vary. Re-calculating: If u_y² = v_y² + 2gy = 4 + 8 = 12, then tanθ = u_y/u_x = √12 / 6 = 2√3/6 = √3/3 = 1/√3 (30°). Checking image: if v_y was larger, answer could be 45°.
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