Homework Help: physics

Posted by Joe on Friday, February 8, 2013 at 7:32pm.

A golfer hits a shot to a green that is elevated 3.00 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 52.0вк 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.

physics - Henry, Saturday, February 9, 2013 at 8:37pm
Vo = 16.1m/s @ 52o.
Yo = 16.1*sin52 = 12.7 m/s.

Y^2 = Yo^2 + 2g*h.
h = (Y^2-Yo^2)/2g=(0-161)/-19.6=8.2 m.

V^2 = Vo^2 + 2g*d.
V^2 = 0 + 19.6*(8.2-3) = 101.92
V = 10.1 m/s.

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