A golfer hits a golf ball from point A with an initial velocity of 50 m/s at an angle of 25° with the horizontal. Determine the radius of curvature of the trajectory described by the ball (a) at point A, (b) at the highest point of the trajectory.
p = V2/an
Vy = V*sin([tex]\theta[/tex])
Vx = V*cos([tex]\theta[/tex])
The Attempt at a Solution
Hello again you wonderful people! Firstly, I would like to say that I only need help with part (a). Part (b) was quite easy as I found that the highest point of trajectory is when the velocities' x component was at its' lowest, and the normal acceleration component acting on it is gravity. (The answer was 209 m!)
My question however, more or less pertains to solving for part a's radius of curvature. My book says that it is 281 m. Yet...I am unsure how they found this. I thought of using kinematics to find some additional information, but there is too much lacking. We are given no time, final velocity, or even starting/ending distances.
Could someone help me out with this problem? I feel like i'm overlooking something silly...again....