HW_3_Solutions_F05 - Physics 112 - HW #3 Solutions 3-16...

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Physics 112 - HW #3 Solutions Fall 2005 Page 1 3-16 [Basic Projectile] (a) To find the time of flight t f , we consider the vertical component of the motion. The time t f to fall from a height h = 1.39 m, with zero vertical velocity initially, is t f = 2 h g = 2(0.75 m) 9.8 m/s 2 = 0.391 s (b) The horizontal component of the velocity v x does not change during the motion. v x can be calculated from the horizontal distance traveled d and the time of flight t f v x = d t f = 1.40 m 0.391 s = 3.58 m/s (c) The vertical component of the velocity just before impact is v y = - g t f = - (9.8 m/s 2 )(0.391 s) = - 3.83 m/s (The negative sign indicates that this velocity component points downwards .) The magnitude of the total velocity is then |v|= v x 2 +v y 2 = 5.24 m/s The direction is θ = arctan |v y | |v x | = 47 ° downwards from horizontal
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Physics 112 - HW #3 Solutions Fall 2005 Page 2 3-22 [Monkey & Zookeeper] Assuming the dart reaches the monkey before it reaches the ground: d = v ox t = (v o cos α o )t t = d v o cos α o The dart reaches the monkey at height: y = v oy t - 1 2 gt 2 = v o sin α o d v o cos α o - 1 2 g d v o cos α o 2 = d tan α o - gd 2 2v o 2 cos 2 α o = (3.00 m) tan42.0° - (9.8 m/s 2 )(3.00 m) 2 2v o 2 cos 2 42.0° = 2.70 m - 79.85 m 3 /s 2 v o 2 (a) y = 2.70 m - 79.85 m 3 /s 2 (12.0 m/s) 2 = 2.70 m - 0.55 m = 2.15 m above the ground (b) y = 2.70 m - 79.85 m 3 /s 2 (8.0 m/s) 2 = 2.70 m - 1.25 m = 1.45 m above the ground (c) y = 2.70 m - 79.85 m 3 /s 2 (4.0 m/s) 2 = 2.70 m - 4.99 m = - 2.29 m Dart hits the ground first. When? y = v oy t - 1 2 gt 2 = 0 t = 2v oy g = 2(4.00 m/s) sin42° 9.8 m/s 2 = 0.546 s Where? x = v ox t = (4.00 m/s)(cos42°)(0.546 s) = 1.62 m (d)
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Physics 112 - HW #3 Solutions Fall 2005 Page 3 3-33 [Circular Motion] For constant speed in a circle, the acceleration is always toward the center of the circle. The magnitude of the acceleration is a = v 2 r = (7.00 m/s) 2 14.0 m = 3.50 m/s 2 . (a)
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HW_3_Solutions_F05 - Physics 112 - HW #3 Solutions 3-16...

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