Question

A golfer hits a shot to a green that is elevated 2.60 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 51.0˚ 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.

Answer 1

consider the motion along the Y-direction

V_oy = initial velocity = 19 Sin51

a = acceleration = - 9.8

Y = vertical displacement = - 2.60

V_fy = final velocity

using the equation

V_fy² = V_oy² + 2 a Y

V_fy² = (19 Sin51)² + 2 (- 9.8) (- 2.60)

V_fy = 16.4 m/s

consider the motion along the X-direction

V_ox = initial velocity = 19 Cos51

a = acceleration = 0 m/s²

using the equation

V_fx = V_ox + a t

V_fx = (19 Cos51) + (0) t

V_fx = 12 m/s

net speed is given as

V_f = sqrt(V_fx² + V_fy²) = sqrt((12)² + (16.4)²) = 20.32 m/s