Department of Natural Sciences Clayton College & State University. Physics 1111 Quiz 3

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1 Clayton College & State University September 16, 2002 Physics 1111 Quiz 3 Name 1. You throw a physics textbook horizontally at a speed of 9.00 m/s from a top of a building. The height of the building is 30.0 m. Ignoring air resistance, a. What are the components of the initial velocity of the book? V 0 : o V 0x = V 0 cos( 0 ) = 9.00m/s cos(0 o ) = 9.00 m/s V 0y = V 0 sin( 0 ) = 9.00m/s sin(0 o ) = 0 m/s b. How long will it take for the book to hit the ground? y 0 = 30.0 m y = 0 m y = y 0 + V 0y t ½ g t 2 0 m = 30.0 m + (0 m/s) t ½ (9.80 m/s 2 )t 2 ½ (9.80 m/s 2 )t 2 = 30.0 m t 2 = 30.0 m/(½ (9.80 m/s 2 )) t = 2.47 s c. How far from the side of the building do you need to walk in horizontal direction to retrieve the textbook? x 0 = 0 m x = x 0 + V 0x t x = 0 m + (9.00 m/s)(2.47 s) x = 22.3 m Clayton State University Physics 1111 Quiz 5.1 September 27, 2005

2 Name SOLUTION 1. You throw a physics textbook with a speed of 9.00 m/s at 25.0 o angle from a top of a building. The textbook lands on the ground 2.50 s later. Ignoring air resistance, V 0 : o a. What are the components of the initial velocity of the book? V x0 = V 0 cos( 0 ) = (9.00 m/s) cos(25.0 o ) = 8.16 m/s V y0 = V 0 sin( 0 ) = (9.00 m/s) sin(25.0 o ) = 3.80 m/s y = y 0 + V y0 t ½ g t 2 0 = y 0 + V y0 t ½ g t 2 y 0 = - V y0 t + ½ g t 2 y 0 = 21.2 m x = x 0 + V x0 t b. How high is the building? c. How far from the side of the building do you need to walk in horizontal direction to retrieve the textbook? x = (0 m) + (8.16 m/s) (2.50 s) =20.4 m Clayton State University Physics 1111 Quiz 4 September 25, 2006 Name SOLUTION 1. A soccer ball is kicked off the 40.0-m cliff at a speed of 20.0 m/s horizontally.

3 a. What are the components of the initial velocity of the ball? V 0 : o V x0 = V 0 cos( 0 ) = (20.0 m/s) cos(0 o ) = 20.0 m/s V y0 = V 0 sin( 0 ) = (20.0 m/s) sin(0 o ) = 0 m/s b. How long is the ball in the air? y = y 0 + V y0 t ½ g t 2 0 = y ½ g t 2 ½ g t 2 = y 0 g t 2 = 2 y 0 t 2 = 2 y 0 /g t = (2 y 0 /g) 1/2 t = (2 (40.0 m) /(9.81 m/s 2 )) 1/2 t = 2.86 s x = x 0 + V x0 t c. How far from the base of the cliff do you need to walk in horizontal direction to retrieve the ball? x = (0 m) + (20.0 m/s) (2.86 s) =57.2 m Clayton College & State University Physics 1111 Quiz 4 February 9, 2004 Name

4 1. A crow is flying horizontally with a constant speed of 2.70 m/s when it releases a clam from its beak. The clam lands on the rocky beach 2.10 s later. Just before the clam lands, what is a. Its horizontal component of velocity? Initial velocity: V 0 : o V 0x = V 0 cos( 0 ) = (2.70 m/s) cos(0 o ) = 2.70 m/s V 0y = V 0 sin( 0 ) = (2.70 m/s) sin(0 o ) = 0 m/s V x = V 0x = 2.70 m/s b. Its vertical component of velocity? V y = V 0y - g t = 0 m/s (9.80 m/s 2 ) (2.10 s) = m/s c. From what height was the clam released? y = 0 m y = y 0 + V 0y t ½ g t 2 0 = y 0 ½ g t 2 y 0 = ½ g t 2 y 0 = 21.6 m d. What is the range of the clam s trajectory? x 0 = 0 x = x 0 + V 0x t x = 0 m + (2.70 m/s) (2.10 s) x = 5.67 m Clayton College & State University Physics 1111 Quiz 4 February 9, 2004

5 Name SOLUTION A military helicopter on a training mission is flying horizontally at a speed of 60.0 m/s and accidentally drops a bomb (fortunately not armed) at an elevation of 300 m. You can ignore air resistance. a. What are the components of the initial velocity? V 0 : o V x0 = V 0 cos( 0 ) = (60.0 m/s) cos(0 o ) = 60.0 m/s V y0 = V 0 sin( 0 ) = (60.0 m/s) sin(0 o ) = 0 m/s b. How much time is required for the bomb to reach the ground? y 0 = 300 m y = 0 y = y 0 + V yo t ½ g t 2 0 = y 0 ½ g t 2 y 0 = ½ g t 2 t 2 = 2 y 0 /g t = (2 y 0 /g) 1/2 t = 7.82 s c. How far does it travel in horizontal direction while falling? x = x 0 + V x0 t x = (0 m) + (60.0 m/s) (7.82 s) = 469 m Clayton State University Physics 1111 Quiz 4 February 7, 2007

6 Name SOLUTION 1. You throw a physics textbook horizontally at a speed of 11.0 m/s from a top of a building. The height of the building is 50.0 m. Ignoring air resistance, V 0 : o a. What are the components of the initial velocity of the book? V x0 = V 0 cos( 0 ) = (11.0 m/s) cos(0 o ) = 11.0 m/s V y0 = V 0 sin( 0 ) = (11.0 m/s) sin(0 o ) = 0 m/s b. How long will it take for the book to hit the ground? y = y 0 + V y0 t ½ g t 2 0 = y ½ g t 2 ½ g t 2 = y 0 g t 2 = 2 y 0 t 2 = 2 y 0 /g t = (2 y 0 /g) 1/2 t = (2 (50.0 m) /(9.81 m/s 2 )) 1/2 t = 3.19 s x = x 0 + V x0 t c. How far from the side of the building do you need to walk in horizontal direction to retrieve the textbook? x = (0 m) + (11.0 m/s) (3.19 s) =35.1 m Clayton State University

7 Physics 1111 Quiz 5 February 13, 2008 Name A projectile is fired with an initial speed of 35.0 m/s at an angle of 45.0 o above the horizontal on a long flat firing range. Determine a. The components of the projectile s initial velocity. b. The time of flight. y = y 0 + V y0 t ½ gt 2 0 = 0 + V y0 t ½ gt 2 0 = t (V y0 ½ gt ) V y0 ½ gt = 0 t = 2 V y0 /g t = 5.05 s V x0 = V 0 cos( 0 ) = (35.0 m/s) cos(45.0 o ) = 24.7 m/s V y0 = V 0 sin( 0 ) = (35.0 m/s) sin(45.0 o ) = 24.7 m/s c. The projectile s range. x = x 0 + V x0 t x = (0 m) + (24.7 m/s) (5.05 s) = 125 m Clayton College & State University Physics 1111 Quiz 5 June 8, 2004 Name SOLUTION

8 A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an angle of 36.9 o above the horizontal. You can ignore air resistance. a. What are the components of the initial velocity? V 0 : o V x0 = V 0 cos( 0 ) = 30.0m/s cos( o ) = 24.0 m/s V y0 = V 0 sin( 0 ) = 30.0m/s sin(36.9 o ) = 18.0 m/s b. The ball is caught at the same elevation as it was hit. How much time was the ball in the air? y = y 0 + V 0y t ½ g t 2 0 = 0 + V 0y t ½ g t 2 0 = t (V 0y ½ g t) (V 0y ½ g t) = 0 2V 0y /g = t t = 3.68 s c. How far did the ball travel in horizontal direction? x = x 0 + V x0 * t x = 0 + (24.0 m/s) * (3.68 s) = 88.2 m d. What is the velocity of the ball at the top of its trajectory? V x = V x0 = 24.0 m/s V y = 0 m/s V: o Clayton State University

9 June 15, 2005 Physics 1111 Quiz 4 Name SOLUTION A famous soccer player, Hedley Footy, kicks the ball at a speed of 15.0 m/s when t = 0 s. The initial velocity vector of the ball makes an angle of 60.0 o with the horizontal direction. a. Find the components of the initial velocity. V 0 : o V x0 = V 0 cos( 0 ) = (15.0 m/s) cos(60.0 o ) = 7.50 m/s V y0 = V 0 sin( 0 ) = (15.0 m/s) sin(60.0 o ) = 13.0 m/s b. If the ball lends at the same level as it started, how long is the ball in motion? y = y 0 + V y0 t - 1/2 g t 2 y = y 0 = 0 0 = V y0 t - 1/2 g t 2 = t ( V y0-1/2 g t) V y0-1/2 g t = 0 2 V y0 / g = t t = 2 (13.0 m/s) / (9.80 m/s 2 ) = 2.65 s x = x 0 + V x0 t c. What is the projectile s range? x = (0 m) + (7.50 m/s)(2.65 s) = 19.9 m V = 7.50 m/s d. What is the ball s speed at the top of trajectory?

10 Remember, V y0 = 0 at the top! Clayton State University June 18, 2007 Physics 1111 Quiz 5 Name SOLUTION A soccer ball is kicked with a speed of 9.50 m/s at an angle of 40.0 o above the horizontal direction. Some time later the ball lands at the same level from which it was kicked. a. Find the components of the ball s initial velocity. V x0 = V 0 cos( 0 ) = (9.50m/s) cos(40.0 o ) = 7.28 m/s V y0 = V 0 sin( 0 ) = (9.50m/s) sin(40.0 o ) = 6.11 m/s b. What are the components of the ball s velocity at the top of the trajectory? V x = V x0 = 7.28 m/s V y = 0 m/s c. How long does it take the ball to reach the top of the trajectory? V y = V y0 gt (V y0 - V y ) /g = t t = (6.11 m/s 0 m/s)/(9.80 m/s 2 ) = s d. How far does the ball go in the horizontal direction? x = x 0 + V x0 ( t) = 0 m + (7.28 m/s) (1.25 s) = 9.07 m

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