Posted by Robert C on Tuesday, December 8, 2009 at 4:41pm.
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]
Related Questions
physics - A golf ball is struck with a five iron on level ground. It lands 92.7 ...
physics - jane hits a golf ball on a flat fairway. The ball leaves the ground at...
Physics - A golf ball is hit with an initial angle of 54.7o with respect to the ...
Physics - A golf ball of mass 30.0 grams is struck by a 4.00 kg golf club for a ...
Physics - A golf ball is given an initial speed of 20 m/s and returns to ground ...
Physics - A golf club contacts a golf ball for 0.15s at an angle to the ...
Physics - A golf ball is hit with an initial angle of 30.5° with respect to ...
physics - A baseball bat makes contact with a ball 1.05 meters above the ground ...
Math/Physics - A golf ball is struck at ground level. The speed of the golf ball...
physics - A ball launched from ground level lands 3 s later on a level field 48 ...
For Further Reading
