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?
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.
You are on the right track! To find the speed of the ball just before it lands, you need to calculate the final horizontal speed (Vx) and the final vertical speed (Vy). Here's how you can do it step by step:
1. Calculate the initial horizontal speed (Vx):
Vx = 17.4 m/s * cos(35.0°)
Vx = 14.2 m/s
2. Calculate the initial vertical speed (Vy):
Vy = 17.4 m/s * sin(35.0°)
Vy = 9.9 m/s
3. Calculate the final vertical speed (Vyf) at the moment the ball lands:
Vyf^2 = Vy^2 - 2 * g * h
Where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height difference between the point where the ball is struck and the elevated green (3.0 m).
Vyf^2 = (9.9 m/s)^2 - 2 * (9.8 m/s^2) * (3.0 m)
Vyf^2 = 98.01 m^2/s^2 - 58.8 m^2/s^2
Vyf^2 = 39.21 m^2/s^2
4. Calculate the final speed of the ball just before it lands (Vf):
Vf = sqrt(Vx^2 + Vyf^2)
Vf = sqrt((14.2 m/s)^2 + (39.21 m^2/s^2))
Vf = sqrt(201.64 + 39.21)
Vf = sqrt(240.85)
Vf = 15.51 m/s
So, the speed of the ball just before it lands is 15.51 m/s.