A golf ball is shot from a tee, elevated 20.0 m above the green, with an initial speed of 27.0 m/s at an angle of 50.0oabove the horizontal. a) What will be the speed of the golf ball just before it hits the green, 20.0 m below where it started? b) How much time will it take for the ball to travel from the tee, where it is shot, to the green? c) What horizontal distance will the ball travel in moving from the shot at the tee to hitting the green?  

Expert Answers

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See the figure below. At point A and B the speed is the same because of the symmetry of the trajectory, only the angle of the speed with the horizontal is reversed.

`V_(By) =-V_(Ay) =-V*sin(alpha) =-27*sin(50) =-20.68 m/s`

`V_(Bx)=V_(Ax) =V*cos(alpha) =-27*cos(50) =17.35 m/s`

At point C the horizontal component of the speed remains the same. On the vertical there is free fall.

`V_(Cx) =V_(Bx) =17.35 m/s`

`V_(Cy) =-sqrt(V_(By)^2 +2*g*h) =-sqrt(20.68^2 +2*9.81*20) =-28.64 m/s`

Thus the total speed at C is

`V_C = sqrt(V_(Cx)^2 +V_(Cy)^2) =sqrt(17.35^2 +28.34^2) =33.23 m/s`

c)

The range `x_0` is

`x_0 =v_0^2/g*sin(2*alpha) =27^2/9.81*sin(2*50) =73.18 m`

The time of movement from B to C is

`V_(Cy) = V_(By) -g*t_1`

`t_1 = (V_(By)-V(Cy))/g = (-20.68+28.64)/9.81 =0.811 s`

The extended range `x_1` is

`x_1 = V_(Bx)*t_1 = 17.35*0.811 =14.08 m`

The total range is

`x = x_0 +x_1 =73.18+14.08 =87.26 m`

b)

The time of movement on the range `x_0` is

`t_0 =x_0/V_(Ax) =73.18/17.35 =4.22 s`

Total time of movement is

`t = t_0 +t_1 =4.22+0.81 =5.03 s`

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