Question
A golfer hits a golf ball, giving it an initial speed of 31.7 m/s at an angle of 55° above the horizontal. How far in meters will the golf ball travel horizontally before hitting the ground?
Answers
Vo = 31.7m/s[55o]
Xo = 31.7*cos55 = 18.18 m/s.
Yo = 31.7*sin55 = 25.97 m/x.
Dx = Vo^2*sin(2A)/g=31.7^2*sin(110)/9.8
= 96.4 m.
Xo = 31.7*cos55 = 18.18 m/s.
Yo = 31.7*sin55 = 25.97 m/x.
Dx = Vo^2*sin(2A)/g=31.7^2*sin(110)/9.8
= 96.4 m.
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what is the value of A in this equation
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