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Published byFerdinand Martin Modified about 1 year ago

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A soccer player kicks a ball into the air at an angle of 34.0° above the horizontal. The initial velocity of the ball is m/s. How long is the soccer ball in the air? = 34.0° Vi = m/s t = ? Vy = Vi*sin t = 2Vy/g Vy = 30.0Sin(34.0) = 16.8 m/s 3.42 s t = -2*16.8/-9.8 = 3.42 s

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What is the horizontal distance traveled by a soccer ball kicked into the air at an angle of 30.0° above the horizontal with initial velocity of m/s? = 30.0° Vi = m/s Dx = ? Vy = Vi*Sin Vx = Vi*Cos t = 2Vy/g Dx = t*Vx Vy = 31Sin(30) = 15.5 Vx = 31Cos(30) = 26.8 t = (-2*15.5)/-9.8 = m Dx = 3.16*15.5 = 84.6 m

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What is the maximum height of a soccer ball kicked into the air at an angle of 32.0° above the horizontal with the initial velocity of m/s? = 32.0° Vi = 25.0 m/s Dy = ? Vy = Vi*Sin Dy = Vy 2 /2g Vy = 25Sin(32) = 13.2 m/s Dy = /(2*9.8) = m

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A ball falls from rest from a height of 452 m. a.How long does it remain in the air? b.If the ball has a horizontal velocity of 1.00*10 2 m/s when it begins to fall, what horizontal displacement will it have? Dy = 452 m Vx = 1.00*10 2 t = ? Dx = ? t 2 = 2Dy/g Dx = t*Vx t 2 = (2*452)/9.8 = √92.2 = 9.60 s Dx = 9.60*100 = 9.60*10 2 m

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An archer stands 40.0 m from the target. If the arrow is shot horizontally with a velocity of 87.0 m/s, how far above the bull’s-eye must she aim to compensate for gravity pulling her arrow downward? Dx = 40.0 m Vx = 87.0 m/s Dy = ? t = Dx/Vx Dy =.5gt 2 t = 40.0/87.0 =.459 s 1.04 m Dy =.5*9.8* = 1.04 m

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A bridge is m above a river. If a lead-weighted fishing line is thrown from the bridge with a horizontal velocity of 24.0 m/s, how far has it moved horizontally when it hits the water? Dy = m Vx = 24.0 m/s Dx = ? t 2 = 2Dy/g Dx = t*Vx t 2 = (2*142.9)/9.8 = √29.2 = 5.40 s Dx = 5.4*24.0 = 1.30*10 2 m

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A beach ball, moving with a speed of m/s, rolls off a pier and hits the water 0.65 m from the end of the pier. How high above the water is the pier? Vx = 1.12 m/s Dx = 0.65 m Dy = ? t = Dx/Vx Dy =.5gt 2 t = 0.65/1.12 =.58 s 1.7 m Dy =.5*9.8*.58 2 = 1.7 m

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A shot put is released with a velocity of 12 m/s and stays in the air for 1.9 s. a.At what angle with the horizontal was it released? b.What horizontal distance did it travel? Vi = 12 m/s t = 1.9 s θ = ? Dx = ? Vy = Vf-at θ = Sin -1 (Vy/Vi) Vx = Vi*Cosθ Dx = Vx*t Vy = 0-9.8*0.95 = 9.3 m/s = Sin -1 (9.3/12) = ° Vx = 12Cos(51) = 7.6 m/s Dx = 7.6*1.9 = m

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A football is kicked at 45° and travels 82 m before hitting the ground. a.What is the initial velocity? b.How long was it in the air? c.How high did it go? = 45° Dx = 82 m Vi = ? t = ? Dy = ? Vi = √(g*Dx)/(2Sin *Cos ) Vx = Vi*Cos t = Dx/Vx Vy = Vi*Sin Dy = Vy 2 /2g 28 m/s Vi = √(9.8*82)/[2Sin(45)Cos(45)] = 28 m/s Vx = 28Cos(45) = 20 m/s 4.1 s t = 82/20 = 4.1 s Vy = 28Sin(45) = 20 m/s 20 m Dy = 20 2 /(2*9.8) = 20 m

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A golf ball is hit with a velocity of 46.0 m/s at 32.0° above the horizontal. Find a.The range of the ball b.The maximum height of the ball Vi = 46.0 m/s = 32.0° Range = Vi 2 Sin(2 )/g Dy = Vy 2 /2g 194 m Range = 46 2 Sin(2*32)/9.8 = 194 m 30.3 m /(2*9.8) = 30.3 m

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