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Starting from Equation (4c) and Equation (4c') and with the conditions in projectile otions (Refer to the module), show that (Vxo)² sin(200) g x = Range = nt: sin(20) = 2 sin(80) cos (80) stify that the range of the projectile is maximum if 80 = 45.0° using this equation.

Starting from Equation (4c) and Equation (4c') and with the conditions in projectile otions (Refer to the module), show that (Vxo)² sin(200) g x = Range = nt: sin(20) = 2 sin(80) cos (80) stify that the range of the projectile is maximum if 80 = 45.0° using this equation.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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2. Starting from Equation (4c) and Equation (4c) and with the conditions in projectile motions (Refer to the module), show that x = Range = (Vxo)² sin(200) g Hint: sin (200) = 2 sin(80) cos (0) Justify that the range of the projectile is maximum if 0o = 45.0° using this equation.
Transcribed Image Text:2. Starting from Equation (4c) and Equation (4c) and with the conditions in projectile motions (Refer to the module), show that x = Range = (Vxo)² sin(200) g Hint: sin (200) = 2 sin(80) cos (0) Justify that the range of the projectile is maximum if 0o = 45.0° using this equation.
(4b) (4b') (4c) (4c') (4d) (4d') X = Xo + ¹/2 (Vxo +Vx) t y = yo + ¹½ (Vyo + Vy) t X = Xo + Vxot + ½ axt² y = yo + Vyot + ½ ayt² vx² = Vxo² + 2ax(x-xo) 2 vy² = Vyo² + 2ay(y-yo)
Transcribed Image Text:(4b) (4b') (4c) (4c') (4d) (4d') X = Xo + ¹/2 (Vxo +Vx) t y = yo + ¹½ (Vyo + Vy) t X = Xo + Vxot + ½ axt² y = yo + Vyot + ½ ayt² vx² = Vxo² + 2ax(x-xo) 2 vy² = Vyo² + 2ay(y-yo)
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