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Unformatted text preview: 4.47.Model:The particle model for the ball and the constantacceleration equations of motion in a plane are
assumed. Visualize: Solve:(a) The time for the ball to fall is calculated as follows: y1 = + ( 1  0 ) + 1 ( 1  0 ) 2 0 0 y 0 = 4 + 0 + 1 ( 9.8 /2 ) ( 1  0 ) 2 Using this result for the horizontal velocity: x1 = 0 + ( 1  0 ) 0 25 m = 0 + 0 ( 0.9035  0 )
2 2 1 = 0.9035 = 27.7 / 0 The friend's pitching speed is 28 m/s. (b) We have v0 y = , where we will use the plus sign for up 5 and the minus sign for down 5. We can 0 write y1 = 0
2 ( 1  0 ) y 0 = 4 2 ( 1  0 )  0  1 0 2 2 1 Let us first find t1 from x1 = 0 + ( 1  0 ) : 0 25 m = 0 + 1 0 Now substituting t1 into the yequation above yields 0m=4 t1 = 25 0 0 25 25  2 0 0 2 2 y = 0 ( 25 )
22 2 4 1 = 22.3 / 44.2 / 25 ) ( The range of speeds is 22 m/s to 44 m/s, which is the same as 50 mph to 92 mph. Assess:These are reasonable speeds for baseball pitchers. 4.53.Model:Use the particle model for the cat and apply the constantacceleration kinematic equations.
Visualize: Solve:The relative velocity of the cat from the mouse's reference frame is (4.0cos301  1.5) /= 1.964 /= 0 30 = (4.0 /) 30 = 2.0 /. The time for the cat to land on the Thus, v0x = = 1.964 / and v0 y = 0 0 floor is found as follows: y1 = + (  0 ) + 1 (  0 ) 2 0 0 1 2 1 0 = 0 + (2.0 /)1 + 1 (9.8 /2 )12 2 y 1 = 0 (trivial solution) and 0.408 s The horizontal distance covered in time t1 is x1 = 0 + (1  0 ) + 1 (1  0 )2 = 0 + (1.964 /)(0.408 ) = 0.802 0 2 That is, the cat should leap when he is 80 cm behind the mouse. ...
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This note was uploaded on 05/20/2008 for the course PHYS 161 taught by Professor Hammer during the Spring '07 term at Maryland.
 Spring '07
 Hammer
 mechanics, Acceleration

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