Posted by Cheryl | Posted in Golf Balls | Posted on 06-11-2009
Tags: balls, golf, math, physics, research
Projectiles launched at an angle: a question of physics?
OK this is my homework problem and I really really really do not understand and am so desperate to please, someone can walk me through this problem??? Does not even matter if you answer or not, I just want to know which formulas to use and how to use … I have lots of problems with this. 1.) A player can hit a golf ball a horizontal distance of over 300 m on a good unit. What maximum height will be a 301.5 m drive reach if it is launched at an angle of 25.0 degrees to earth? (Hint: In the top of its flight, the vertical component of the ball speed is 0) thanks sooooooooooooooo much for anyone who helps me
d = v ^ 2 Sin (2 @) / g, where @ is the angle and v is the initial velocity, so 301.5 = v ^ 2 sin (2 * 25) / 9.8 v = 62.1054 m / s vy = v sin @ = 62.1054 * sin 25 = 26.2469 m / s eqaution kinematics (VYF ) ^ 2 = (vyo) ^ 2 +2 (y) (h) for 0 = 26.2469 ^ 2-2 (9.8) (h) h = 35.1479 m
