PS3-7Solve with complete step-by-step process / solution. 2. Starting from Equation (4c) and Equation (4c) and with the conditions in projectilemotions (Refer to the module), show thatx = Range =(Vxo)² sin(200)gHint: 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) ty = yo + ¹½ (Vyo + Vy) tX = Xo + Vxot + ½ axt²y = yo + Vyot + ½ ayt²vx² = Vxo² + 2ax(x-xo)2vy² = Vyo² + 2ay(y-yo)

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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

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter33: Particle Physics

Section: Chapter Questions

Problem 25CQ: The sigmazero particle decays mostly via the reaction Explain how this decay and the respective...

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Question

PS3-7

Solve with complete step-by-step process / solution.

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)

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