To find the speed of the ball just before it lands, we have to consider the projectile motion of the ball. The initial velocity of the ball is given by the speed and the angle of projection. We have, v = 19.0 m/s and θ = 54.0°. The horizontal and vertical components of velocity are given byv_x = v cos θ and v_y = v sin θ.From the equation of motion, we know that the time of flight (t) of the projectile is given byt = 2v sin θ/gwhere g is the acceleration due to gravity (9.8 m/s²).The maximum height reached by the projectile can be found by using the formula:h = v² sin² θ/2gFrom the conservation of energy, the final kinetic energy (K) of the projectile just before it lands is equal to the initial kinetic energy (K₀) plus the potential energy (U) due to its height above the ground. We have,K + U = K₀ => ½ mv² + mgh = ½ mv₀²where m is the mass of the ball, h is the height of the green, and v₀ is the initial speed of the ball. Since the ball is not given to be rolling after the shot, the final velocity of the ball is v.Finally, we can use the horizontal component of velocity to find the distance (D) covered by the ball before landing. We have,D = v_x tSo, we can calculate the speed (v) of the ball just before it lands by using the following steps:Calculate the horizontal and vertical components of velocity using the given values of speed and angle of projection.Calculate the time of flight and the maximum height reached by the projectile using the formulas above.Use the conservation of energy to find the initial speed of the ball.Use the horizontal component of velocity and the time of flight to find the distance covered by the ball before landing.Use the distance and time of flight to find the speed of the ball just before it lands. Answer: 16.7 m/s.