# 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 50.0o above the horizontal. a) What will be the speed of the golf ball just before...

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 50.0o

above 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?

### 1 Answer | Add Yours

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.