91_PartUniversity Physics Solution

91_PartUniversity - Motion in Two or Three Dimensions 3-21 SET UP Take the origin of coordinates at the point where the canister is released Take y

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Motion in Two or Three Dimensions 3-21 SET UP: Take the origin of coordinates at the point where the canister is released. Take y + to be upward. The initial velocity of the canister is the velocity of the plane, 64.0 m/s in the -direction. x + Figure 3.53 Use the vertical motion to find the time of fall: ?, t = 0 0, y v = 2 9.80 m/s , y a =− 0 90.0 m yy −= (When the canister reaches the ground it is 90.0 m below the origin.) 2 1 00 2 y y yy v t a t −= + EXECUTE: Since 0 0, y v = 0 2 2( ) 2( 90.0 m) 4.286 s. 9.80 m/s y t a −− == = SET UP: Then use the horizontal component of the motion to calculate how far the canister falls in this time: 0 ?, xx 0, x a 0 64.0 m/s, x v = EXECUTE: 2 1 2 (64.0 m/s)(4.286 s) 0 274 m. xx v t = += EVALUATE: The time it takes the cannister to fall 90.0 m, starting from rest, is the time it travels horizontally at constant speed. 3.54. IDENTIFY: The equipment moves in projectile motion. The distance D is the horizontal range of the equipment plus the distance the ship moves while the equipment is in the air. SET UP: For the motion of the equipment take x + to be to the right and y + to be upwards. Then 0 x a = , 2 9.80 m/s y a , 0 cos 7.50 m/s x vv α and 0 sin 13.0 m/s y . When the equipment lands in the front of the ship, 0 8.75 m . EXECUTE: Use the vertical motion of the equipment to find its time in the air: 2 1 2 y y t gives ( ) 2 1 13.0 ( 13.0) 4(4.90)(8.75) s 9.80 t + . The positive root is 3.21 s t = . The horizontal range of the equipment is 2 1 2 (7.50 m/s)(3.21 s) 24.1 m t = = . In 3.21 s the ship moves a horizontal distance (0.450 m/s)(3.21 s) 1.44 m = , so 24.1 m 1.44 m 25.5 m D =+= . EVALUATE: The equation 2 sin2 v R g = from Example 3.8 can’t be used because the starting and ending points of the projectile motion are at different heights. 3.55. IDENTIFY: Projectile motion problem. Take the origin of coordinates at the point where the ball leaves the bat, and take y + to be upward. 0 cos x = 0 sin , y = but we don&t know 0 . v Figure 3.55 Write down the equation for the horizontal displacement when the ball hits the ground and the corresponding equation for the vertical displacement. The time t is the same for both components, so this will give us two equations in two unknowns ( 0 v and t ).
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3-22 Chapter 3 (a) SET UP: y -component : 2 9.80 m/s , y a =− 0 0.9 m, yy −= 00 sin45 y vv = ° 2 1 2 yy v ta t −= + EXECUTE: 22 1 0 2 0.9 m ( sin45 ) ( 9.80 m/s ) vt t ° + SET UP: x -component : 0, x a = 0 188 m, xx cos45 x 2 1 2 x x x xv t a t EXECUTE: 0 188 m cos45 x t == ° Put the expression for t from the x -component motion into the y -component equation and solve for 0 . v (Note that cos45 . °= ° ) 2 2 0 188 m 188 m 0.9 m (4.90 m/s ) cos45 cos45 v ⎛⎞ ° ⎜⎟ °° ⎝⎠ 2 2 0 188 m 4.90 m/s 188 m 0.9 m 188.9 m cos45 v =+= ° 2 2 0 cos45 4.90 m/s , 188 m 188.9 m v ° = 2 0 188 m 4.90 m/s 42.8 m/s cos45 188.9 m v ° (b) Use the horizontal motion to find the time it takes the ball to reach the fence: SET UP: x -component : 0 116 m, 0, x a = cos45 (42.8 m/s)cos45 30.3 m/s,
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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91_PartUniversity - Motion in Two or Three Dimensions 3-21 SET UP Take the origin of coordinates at the point where the canister is released Take y

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