Question

A golfer hits a shot to a green that is elevated 3.10 m above the point where the ball is struck. The ball leaves the club at a speed of 19.4 m/s at an angle of 33.0˚ 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.

I would like a clear explanation for this. Ive tried it 3 time and have gotten different answers each time.
7.89
8.6
12.67

I only have one more try at this question.

Answer 1

Vo = 19.4m/s @ 33o
Xo = 19.4*Cos33 = 16.27 m/s
Yo = 19.4*sin33 = 10.57 m/s.

Y^2 = Yo^2 + 2g*h = 0 @ max ht.
h = -(Yo^2)/2g = -(10.57^2)/-19.6 = 5.7m
Above gnd.

Y^2 = Yo^2 + 2g*d
Y^2 = 0 + 19.6*(5.7-3.1) = 50.96
Y = 7.14 m/s. = Ver. component of velocity.

V^2 = Xo^2 + Y^2 = 16.27^2 + 7.14^2 =
315.69
V = 17.8 m/s. = Total velocity.