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 16.1 m/s at an angle of 52.0¢ª 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

Vo = 16.1m/s @ 52o.
Yo = 16.1*sin52 = 12.7 m/s.

Y^2 = Yo^2 + 2g*h.
h = (Y^2-Yo^2)/2g=(0-161)/-19.6=8.2 m.

V^2 = Vo^2 + 2g*d.
V^2 = 0 + 19.6*(8.2-3) = 101.92
V = 10.1 m/s.