A golfer hits a shot to a green that is elevated 3.20 m above the point where the ball is struck. The ball leaves the club at a speed of 19.0 m/s at an angle of 52.0˚ 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.

Question

v = Type your answer here Choose your answer here

Answer 1

initial velocity components
... vertical = 19.0 m/s * sin(52.0º)
... horizontal = 19.0 m/s * cos(52.0º)

final velocity components
... fvertical = √{[19.0 m/s * sin(52.0º)]^2 - (2 * g * 3.20 m)}
... fhorizontal = 19.0 m/s * cos(52.0º)

final velocity = √[(fvertical ^ 2) + (fhorizontal ^ 2)]