Class Management Help 04 Kinematics in 2 Dimensions Begin Date: 9/11/2017 8:00:0

Question

Class Management Help 04 Kinematics in 2 Dimensions Begin Date: 9/11/2017 8:00:00 AM.. Due Date: 9/18/2017 7:50:00 AM End Date: 9/22/2017 7:50:00 AM (11%) Problem 3: A baseball slugger snacks a pitch and watches the ball drift into the bleachers for a home run, landing h 8.5 m higher than it was struck. When visiting with the fan that caught the ball, he learned the ball was moving with final velocity vr= 39, 45 m/s at an angle ef= 26,5° below horizontal when caught. Assume the ball encountered no air resistance, and use a Cartesian coordinate system with the origin located at the ball's initial position. 0, Otheexpertta.com 25% Part (a) Create an expression for the ball's initial horizontal velocity, vo , in terms of the variables given in the problem statement. Grade Summary Deductions Potential 100% Submissions Attempts remaining: & GRe per attempt) sin() sin(o) sinep) detailed view Submit Hnt I give up Hints:dediction per hant Hints remaining- Feedback:-% deduction per feedback. 25% Part (b) Calculate the ball's initial vertical velocity, vo , in ms. 25% Part (c) Calculate the magnitude of the ball's initial velocity, va in ms. 25% Part (d) Find the angle 60 in degrees above the horizontal at which the ball left the bat. All cosben @ 30Y Eapen TA LLC

Explanation / Answer

a)V0x = Vf cos(theta)

b)V0x = Vf cos(theta) = 39.45 x cos26.5 = 35.31 m/s

tan(theta) = Vy/Vx => Vy = Vx tan(theta)

Vy = 35.31 x tan(-26.5) = -17.61 m/s

V0y = sqrt (Vy^2 + 2 a S)

Voy = sqrt (-17.61^2 + 2 x 9.8 x 8.5) = 21.83 m/s

Hence, V0y = 21.83 m/s

c)V0 = Voy/sin(theta) = 21.83/sin26.5 = 48.93 m/s

Hence, V0 = 48.93 m/s

d)theta = tan- (21.83/35.31) = 31.73 deg

Hence, theta = 31.73 deg

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