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gun, Caiculate no W Tai uiC II IS away nomT uc guu (0.04A1TO I) 3. A ball is fired from a cannon inclined at 36.9° above the horizontal. The speed with which it leaves the barrel is 200 m/s and the mouth of the cannon is 2.00 m above ground level. Assume level ground. What are the horizontal and vertical components of the initial velocity of the cannon ball? Determine the velocity of the cannon ball at t = 6.0 s. What is the ball's velocity when it reaches its maximum height? How far down range (from the cannon mouth) does the ball land? (b) (d)

gun, Caiculate no W Tai uiC II IS away nomT uc guu (0.04A1TO I) 3. A ball is fired from a cannon inclined at 36.9° above the horizontal. The speed with which it leaves the barrel is 200 m/s and the mouth of the cannon is 2.00 m above ground level. Assume level ground. What are the horizontal and vertical components of the initial velocity of the cannon ball? Determine the velocity of the cannon ball at t = 6.0 s. What is the ball's velocity when it reaches its maximum height? How far down range (from the cannon mouth) does the ball land? (b) (d)

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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Please answer question and just send me the paper solutions asap dont type the answer question 3abcd only And use the formula that is given dont use any other formula and use 9.81 for acceleration not 9.8 if required please solve it faster asap and dont give wrong answer
Equations and Constants F = µPm sinA sinB sinc Fnet = ma a? = b2 + c? - 2bccosA a b. C V2 = V1 + aAt Ad Ad = v;At + ½aAť Ad = v2At - ½aAť Vav Ad = %(v1 + v2)At (vz) = (v1} + 2aAd v2 a. = g = 9.81 4n rf2 T2 W FAd W= ΔΕ PE = mgAy 1 KE = mv2 2. F = kx Elastic Energy kx2 P= mv FAt = Ap GMm mearth = 5.972 x 1024kg mmoon = 7.348 x 1022kg Fg Nm2 G = 6.67x10-11 kg2 re = 6378 km Tmoon = 1737 m kq,92 F. = mproton = 1.67 x 10-27kg melectron = 9.11 x 10-31kg Nm2 k = 9.00x109 C2 lelectron/proton +1.60x10-19C Fmagnetic = qvBsin®
Transcribed Image Text:Equations and Constants F = µPm sinA sinB sinc Fnet = ma a? = b2 + c? - 2bccosA a b. C V2 = V1 + aAt Ad Ad = v;At + ½aAť Ad = v2At - ½aAť Vav Ad = %(v1 + v2)At (vz) = (v1} + 2aAd v2 a. = g = 9.81 4n rf2 T2 W FAd W= ΔΕ PE = mgAy 1 KE = mv2 2. F = kx Elastic Energy kx2 P= mv FAt = Ap GMm mearth = 5.972 x 1024kg mmoon = 7.348 x 1022kg Fg Nm2 G = 6.67x10-11 kg2 re = 6378 km Tmoon = 1737 m kq,92 F. = mproton = 1.67 x 10-27kg melectron = 9.11 x 10-31kg Nm2 k = 9.00x109 C2 lelectron/proton +1.60x10-19C Fmagnetic = qvBsin®
gun, Caiculatc no W Tai uiT II IS away HO uc guuT (0.04A1TOʻ III) 3. A ball is fired from a cannon inclined at 36.9° above the horizontal. The speed with which it leaves the barrel is 200 m/s and the mouth of the cannon is 2.00 m above ground level. Assume level ground. What are the horizontal and vertical components of the initial velocity of the cannon ball? Determine the velocity of the cannon ball at t = 6.0 s. What is the ball's velocity when it reaches its maximum height? How far down range (from the cannon mouth) does the ball land? (b) (d)
Transcribed Image Text:gun, Caiculatc no W Tai uiT II IS away HO uc guuT (0.04A1TOʻ III) 3. A ball is fired from a cannon inclined at 36.9° above the horizontal. The speed with which it leaves the barrel is 200 m/s and the mouth of the cannon is 2.00 m above ground level. Assume level ground. What are the horizontal and vertical components of the initial velocity of the cannon ball? Determine the velocity of the cannon ball at t = 6.0 s. What is the ball's velocity when it reaches its maximum height? How far down range (from the cannon mouth) does the ball land? (b) (d)
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