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`

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.