#2. A football kicker is attempting a field goal from 45 mout. The ball is kicke
ID: 1759956 • Letter: #
Question
#2. A football kicker is attempting a field goal from 45 mout. The ball is kicked and just clears the bar 3.6s later. If theheight of the bar is 3.0 m: a) what was the initial speed of the ball? b)what is the maximum height above ground of the ball duringits fleight? c) what is the angle the ball was kicked above thehorizontal? #2. A football kicker is attempting a field goal from 45 mout. The ball is kicked and just clears the bar 3.6s later. If theheight of the bar is 3.0 m: a) what was the initial speed of the ball? b)what is the maximum height above ground of the ball duringits fleight? c) what is the angle the ball was kicked above thehorizontal?Explanation / Answer
You have to write the motion equation for bothdirections: . final position = initialposition + init velocity * time + (1/2) at2 . horizontal: 45 = 0 + v cos *3.6 + 0 . vertical: 3 = 0 + v sin * 3.6 - (1/2) * 9.8 *3.62 . Now you have two equations with two unknowns (angle the ballwas kicked and the initial speed). Solve for each of these, whichare the answers to parts a and c: . 12.5 = vcos 18.47 = v sin . tan = 18.47 /12.5 = 55.92 degrees . v = 12.5 /cos55.92 = 22.305 m/s . And max height = (vsin)2 / 2g = (18.47)2 / 2* 9.8 = 17.41 meters . So the three answers: . (a) 22.305m/s (b) 17.41m (c) 55.92degrees . Now you have two equations with two unknowns (angle the ballwas kicked and the initial speed). Solve for each of these, whichare the answers to parts a and c: . 12.5 = vcos 18.47 = v sin . tan = 18.47 /12.5 = 55.92 degrees . v = 12.5 /cos55.92 = 22.305 m/s . And max height = (vsin)2 / 2g = (18.47)2 / 2* 9.8 = 17.41 meters . So the three answers: . (a) 22.305m/s (b) 17.41m (c) 55.92degreesRelated Questions
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