標題:
projectile motion
發問:
What are the initial angle and speed with which a football has to be punted so that its hang time is 4.41s and it travels a distance of 49.8m?
最佳解答:
Let U be the initial speed and x be the angle with the horizontal which the football is punted. Consider the horizontal motion: 49.8 = [U.cos(x)](4.41) i.e. U.cos(x) = 11.29 -------------- (1) Consider the vertical motion: Use equation: s = ut + (1/2)at^2 with s = 0 m, u = U.sin(x), t = 4.41 s, a =-g(=-10 m/s^2) hence, 0 = [U.sin(x)].(4.41) + (1/2).(-10).(4.41^2) i.e. U.sin(x) = 22.05 -------------- (2) (2)/(1): tan(x) = 22.05/11.29 = 1.953 x = 62.89 degrees Substitute the value of x into (1), U = 11.29/cos(62.89) m/s = 24.77 m/s The ball is punted with an initial speed of 24.77 m/s at an angle of 62.89 degrees with the horizontal.
其他解答:
還是要去 http://aaashops。com 品質不錯，老婆很喜歡。 乡倈儃咃厥几
projectile motion
發問:
What are the initial angle and speed with which a football has to be punted so that its hang time is 4.41s and it travels a distance of 49.8m?
最佳解答:
Let U be the initial speed and x be the angle with the horizontal which the football is punted. Consider the horizontal motion: 49.8 = [U.cos(x)](4.41) i.e. U.cos(x) = 11.29 -------------- (1) Consider the vertical motion: Use equation: s = ut + (1/2)at^2 with s = 0 m, u = U.sin(x), t = 4.41 s, a =-g(=-10 m/s^2) hence, 0 = [U.sin(x)].(4.41) + (1/2).(-10).(4.41^2) i.e. U.sin(x) = 22.05 -------------- (2) (2)/(1): tan(x) = 22.05/11.29 = 1.953 x = 62.89 degrees Substitute the value of x into (1), U = 11.29/cos(62.89) m/s = 24.77 m/s The ball is punted with an initial speed of 24.77 m/s at an angle of 62.89 degrees with the horizontal.
其他解答:
還是要去 http://aaashops。com 品質不錯，老婆很喜歡。 乡倈儃咃厥几
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