A golfer hits a shot to a green that is elevated 3 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 41.1° 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

original component of velocity up = Vi
= 16.1 sin 41.1 = 10.6 m/s

H = Hi + Vi t - (1/2) 9.81 t^2
3 = 0 + 10.6 t - 4.9 t^2
so
4.9 t^2 - 10.6 t + 3 = 0
t^2 - 2.16 t + .612 = 0
t = [ 2.16 +/- sqrt(4.67-2.44) ]/2
t = 1.08 +/- .75
t = 1.83 seconds or .33 seconds
Use the 1.83 seconds. The .33 is on the way up just after you hit it.
What is the speed at t = 1.83
horizontal speed is 16.1 cos 41.1
= 12.1 m/s
vertical speed:
v = Vi -9.81 t
v = 10.6 - 9.81(1.83)
v = -7.35 m/s
total speed = sqrt (12.1^2 + 7.35^2)
= 14.15 m/s

Answer 2

your working is amazing dude