posted by Robert C .
A golf ball is struck with a five iron on level ground. It lands 85.0 m away 4.60 s later. What was the magnitude and direction of the initial velocity
It depends upon the initial "launch angle" of the golf ball, and the "launch velocity". The time spend coming down before it hit the ground was 2.3 s. The vertical velocity component was therefore Vyo = g*2.3s = 22.56 m/s. The horizontal velcoity component (which remains constant) is Vx = 85m/4.6s = 18.48 m/s. The "launch angle" is arctan 22.56/18.48 = 50.7 degrees
The initial velcoity is sqrt[(Vyo^2 + Vx^2]