Question

A golfer hits a shot to a green that is elevated 3.00 m above the point where the ball is struck. The ball leaves the club at a speed of 17.1 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.

-- the answer is not 11.07 m/s --

Answer 1

Resolving velocity into components, Horizontal is u_x = 17.1 cos52 = 10.53 m/s.

Vertical is u_y = 17.1 sin52= 13.47 m/sec.

Maximum height reached =H= u_y² /2g = 9.26 m.

From the maximum point, it falls on green at a height of 3m. hence we can understand that it travels a height (H-3)m from the highest point.

At maximum height, u_y becomes 0. Under action of gravity it again falls a distance (H-3) = 6.26m.

Hence it becomes, V_y² - 0 = 2g(H-3)

                          V_y = 11.08 m/sec.

Horizontal component remains unchanged. Hence V_x = 10.53 m/s.

Total velocity just before it hits the green = V = ?(V_x²+V_y²)= 15.29 m/sec