Question

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 14.2 m/s at an angle of 43.2° 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 = 14.2m/s[43.2o]
Yo = 14.2*sin43.2 = 9.72 m/s.

hmax = (Y^2-Yo^2)/2g
hmax = (0-9.72^2)/19.6 = 4.82 m.

Y^2 = Yo^2 + 19.6*h
Y^2 = 9.72^2 + 19.6*(4.82-3) = 130.15
Y = 11.41 m/s

Answer 2

CORRECTION:
Y^2 = 0 + 19.6*(4.82-3) = 35.67
Y = 5.97 m/s.

Answer 3

that answer is wrong

Answer 4

then correct it