Calculate Golf Ball Velocity Components: x & y

[DarkTangent]

A golf ball leaves an 8 iron at an angle of 35.0 degrees at a velocity of 86.4 m/s. Determine the horizontal [x] and vertical [y] components of the velocity.


There really were no equations for this, I attempted to use my protractor and ruler for this problem and ended up with a triangle. I used sin and cos to find my components, but I'm not so certain.



I got 49.6 m/s for the y-component and 70.8 m/s for the x-component. Any idea if those are right?

 

[neutrino]

Yes, those are right. (approximately, of course)

 

[m2003]

I would recommend using the equations for projectile motion to calculate the horizontal and vertical components of the golf ball's velocity. The horizontal component (x) can be found using the equation Vx = V*cosθ, where V is the initial velocity and θ is the angle of launch. In this case, Vx = 86.4*cos35.0 = 70.8 m/s. The vertical component (y) can be found using the equation Vy = V*sinθ - gt, where g is the acceleration due to gravity and t is the time the ball spends in the air. Since the ball is launched horizontally, there is no initial vertical velocity, so Vy = -gt. Using the value of g (9.8 m/s^2) and the time the ball spends in the air (which can be calculated using the equation t = 2V*sinθ/g), we get Vy = -9.8*(2*86.4*sin35.0/9.8) = -49.6 m/s. Therefore, the y-component of the velocity is -49.6 m/s, indicating that the ball is moving downward. Your calculations for the x and y components of the velocity seem to be correct. However, I would recommend using the equations for more accurate results.