1₁ A projectile is fired from ground level with an initial Speed of 55.6m/s at an angle of 41. 2° above the horizantal. Neglect air resistance, take upward as the postive direction, and use g = 9.80 m/s ². 2 (a) Determine the max height reached by the projectile. > V = 55.6m/s

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Chapter1: Units, Trigonometry. And Vectors
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A projectile is fired from ground level with an initial
Speed of 55.6m/s at an angle of 41. 2° above the horizantal.
Neglect air resistance, take upward as the postive
direction, and use g = 9.80 m/s ².
(a) Determine the max height reached by the projectile.
V₁-55.6m/s
V
41.20
N
41.2
Vix
Viy
Viy = Visin - (55-6m/s) sin 41.21
= 36.62
= 36.6 m/s
Vix = 55.6 m/s cos(41.29)
= 41, 8 m/s
Vya Viy + by th
ay
m/s +
Transcribed Image Text:A projectile is fired from ground level with an initial Speed of 55.6m/s at an angle of 41. 2° above the horizantal. Neglect air resistance, take upward as the postive direction, and use g = 9.80 m/s ². (a) Determine the max height reached by the projectile. V₁-55.6m/s V 41.20 N 41.2 Vix Viy Viy = Visin - (55-6m/s) sin 41.21 = 36.62 = 36.6 m/s Vix = 55.6 m/s cos(41.29) = 41, 8 m/s Vya Viy + by th ay m/s +
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