Asked by Carissa
I'm working on projectile motion and I am supposed to find the speed for this problem, however I'm having trouble setting up the problem. I'm not finding any help from the book either.
A golfer hits a shot to a green that is elevated 2.50 m above the point where the ball is struck. The ball leaves the club at a speed of 18.7 m/s at an angle of 51.0˚ above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
A golfer hits a shot to a green that is elevated 2.50 m above the point where the ball is struck. The ball leaves the club at a speed of 18.7 m/s at an angle of 51.0˚ above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
Answers
Answered by
Henry
Vo = 18.7m/s[51o]
Xo = 18.7*Cos51 = 11.8 m/s.
Yo = 18.7*sin51 = 14.5 m/s.
Y^2 = Yo^2 + 2g*h = 0
h = -Yo^2/2g = -14.5^2/-19.6 = 10.8 m.
Y^2 = Yo^2 + 2g*d = 0 + 19.6*(10.8-2.5)= 162.68
Y = 12.8 m/s = Ver. component.
V^2 = Xo^2 + Y^2 = 11.8^2 + 12.8^2 = 303.08
V = 17.4 m/s. = Speed just before landing.
Xo = 18.7*Cos51 = 11.8 m/s.
Yo = 18.7*sin51 = 14.5 m/s.
Y^2 = Yo^2 + 2g*h = 0
h = -Yo^2/2g = -14.5^2/-19.6 = 10.8 m.
Y^2 = Yo^2 + 2g*d = 0 + 19.6*(10.8-2.5)= 162.68
Y = 12.8 m/s = Ver. component.
V^2 = Xo^2 + Y^2 = 11.8^2 + 12.8^2 = 303.08
V = 17.4 m/s. = Speed just before landing.
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