physics
posted by Dee on .
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.

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(4.672.44) ]/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