The trajectory of the ball is symmetric with respect to the highest point (let it be the point A). Point B (the point where the ball hits the green), point C is at the rising branch of trajectory, and the initial point O.
The magnitudes of velocities at the points B and C are equal to each (they differ only in directions)
v(x) =v(ox) =v(o) •cosα = 17.4•cos35 = 14.3 m/s.
v(oy) = v(o) •sinα = 17.4•sin35 =10 m/s.
h = {v(oy)² - v(y)²}/2g.
v(y) =sqrt(v(oy)² -2gh) =
=sqrt{10² - 1•9.8•3)=8.4 m/s.
v =sqrt(v(x)² v(y)²} = sqrt{ 14.3² + 8.4² ) =21.9 m/s.
A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club at a speed of 17.4 m/s at an angle of 35.0¢ª 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.
I was going to do vx= 17.4 cos 35
vy=(17.4 sin 35) - 9.8(?)
and then do sqrt vx^2 + vy^2 but I wasn't sure if the vy equation was correct?
Physics(Please respond) - drwls, Monday, May 28, 2012 at 8:50pm
Vx remains constant. You are correct on its value.
Vy2 decreases to Vyo^2 - 2 g *(3 m)
Then take the sqrt of Vx^2. + Vy^2
so for vy I would do (17.4 sin 35)^2 - 2(9.8)(3) ?? And for vx= 17.4cos35?
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