A golfer hits a shot to a green that is elevated 2.70 m above the point where the ball is struck. The ball leaves the club at a speed of 19.0 m/s at an angle of 36.0˚ 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.

1 answer

Vo = 19m/s[36o].
Xo = 19*Cos36 = 15.4 m/s.
Yo = 19*sin36 = 11.2 m/s.

Y^2 = Yo^2 + 2g*h = 0.
h = -Yo^2/2g = -(11.2^2)/-19.6 = 6.4 m.

Y^2 = Yo^2 + 2g*h = 0 + 19.6(6.4-2.7) =
72.52.
Y = 8.5 m/s = Ver. component of final
velocity.

V = sqrt(Xo^2+Y^2) = sqrt(15.4^2+8.5^2)= 17.6 m/s.