a) the vertical componant of the velocity at launch is given by

`V_(y_(o)) = Vsin(theta) = 27.0 m/s sin(50) = 20.693 m/s`

The problem is symetrical so on the way back down, when the ball reaches the height of the tee it will have a vertical velocity of -20.693 m/s. As it continues to fall the additional 20.0 meters it will reach a final vertical velocity given by

`V_(y) =sqrt(V_(y_(o))^2 + 2*a*d) = 28.639 m/s down`

The final speed (velocity without direction) is given by

`V = sqrt(V_(y)^2 + V_(x)^2)` where `V_(x) = Vcos(theta) = 17.355 m/s`

`V = 33.5 m/s` You could also get the angle to determine the final velocity from `tan^(-1)(V_(y)/V_(x)) =-57^o` or 57 degrees below the horizon.

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