<p>For Cutnell Physics 8th Edition Chapter 3 Problem 20P<br /><br />A golfer hit
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Question
<p>For Cutnell Physics 8th Edition Chapter 3 Problem 20P<br /><br />A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club at a speed of 14.0 m/s^2 at an angle of 40 degrees above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball right before it lands. <br /><br />The solution for this on Cramster looked wrong, and I couldn't understand the logic of finding the right formulas. I can find Initial velocity for x and y, and I need to take that and get it into the final velocity. </p>Explanation / Answer
A golfer hits a shot to a green that is elevated h =3.0 m above the point where the ball is struck. The ball leaves the club at a speed of u= 14.0 m/s angle =400 ______________________________________________________________________________ the maximum height ball can reach is H=(usin)^2/2g =(14sin400)^2/2*9.8 =4.13 m ball down by the distance is h =H -3 =1.13 m the vertical velocity of the ball is vy=2gh =2*9.8*1.13 =4.7 m/s horizontal velocity vx=ucos =14cos400 =10.72 m/s the resultant velocity of the ball is v = vx^2+ vy^2 =11.70 m/s the final velocity of the ball is v =11.70 m/s =10.72 m/s the resultant velocity of the ball is v = vx^2+ vy^2 =11.70 m/s the final velocity of the ball is v =11.70 m/sRelated Questions
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