Projective Motion Worksheet Answers PDF

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Projective Motion Worksheet Answers...

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Pd

Par t i cl eModel si nTwoDi mensi ons:

Pr oj ect i l eMot i onRevi ew 1. A soccer goalie makes a save and then kicks the ball through the air to the middle of the field. c. Graph the vertical component of a. Graph the horizontal component the ball's motion while in the air. of the ball's motion while in the air. d. Explain what each graph shows in b. Explain what each graph shows in This graph shows the ball is moving in the positive direction words. words. and slowing down for the first half of the trip. For the second half of the trip it is moving in the negative direction and speeding up. The constant slope shows the ball is moving in the positive direction at a constant velocity.

This graph shows the ball’s velocity is always changing. The constant slope shows the ball has a constant negative acceleration.

This graph shows the ball is moving in the positive direction at a constant velocity. The horizontal slope shows the ball is not accelerating.

This graph shows the ball has a constant This graphacceleration. shows the ball is not negative accelerating.

e. Draw a force diagram for the soccer ball while it is in the air. Fg vy

vy v a vy motion f. Draw map for the soccer ball's trajectory. v v v

vy v

vx vx

vx

vx

vx vy v

vx vy v

a a

vx a

vx ©Modeling Instruction 2013

vy v

a

a a a

vx 1

a

U6 2D Motion - review v3.1 a

©Modeling Instruction 2013

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U6 2D Motion - review v3.1

2. Tom the cat is chasing Jerry the mouse across a table surface 1.5 m high. Jerry steps out of the way at the last second, and Tom slides off the edge of the table at a speed of 5 m/s. a. Where will Tom strike the floor? b. What are his vertical and horizontal velocity components just before he hits the floor?  y v yi t  12 a t 2  t 

2y 2(-1.5m )   0.55s a -10 sm2

 x v x t   x 5 ms (0.55s)  2.7m

y x y = -1.5m vyi = 0 m a = g = -10 s 2 t = ? x = ?

The horizontal velocity is always 5.0

m s

.

vyf vyi  at 0 (-10 sm2 ).55 s  5.5 ms

3. A lacrosse player slings the ball at an angle of 30 degrees above the horizontal with a speed of 20 m/s. How far away should a teammate position herself to catch the ball? Assuming toss and catch at same height off the ground. v = 20 30∞

vx

y = 0m vyi = 10 vx = 17.3 a = g = -10 t = ? x = ?

vyi vx  cos30 (20 ms ) 17.3 ms vyi sin30 (20 ms ) 10 ms y v yi t  12 a t 2  0  (10 ms )t  12 ( 10 sm2 )t 2 (5 sm2 )t 10 ms  t  2.0s x v x t (17.3 ms )2.0s 35 m

4.

A ball is thrown straight upward and returns to the thrower's hand after 3 seconds in the air. A second ball is thrown at an angle of 30 degrees with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically? y = 0m m a = g = -10 s 2 t = 3.0s vyi = 10 ms v=?

v = 15

yi ©Modeling Instruction 2013 30∞

vx

 y v yi t  12 a t 2  0 v yi (3.0s )  12 ( 10 sm2 )(3.0s )2 vyi (3.0s ) (5.0 ms 2 )(3.0s)2  vyi (5.0 ms 2 )(3.0s ) 15 ms For a ball thrown at an angle to get to the same height, it must have the same initial vertical velocity. 3

U6 2D Motion - review v3.1

sin30 

©Modeling Instruction 2013

15 ms 15 ms  v  30 ms  sin30 v

4

U6 2D Motion - review v3.1

5. Two objects are initially the same height above the ground. Simultaneously, one is released from rest and the other is shot off horizontally with an initial speed of 2.5 m/s. The two objects collide after falling 20 m. How far apart were the objects initially? y = -20 m vyi = 0

y

vx = 0 a = g = -10 t = ? x = 0

x 2 y 2(-20m )  2.0s a -10 ms2

y = -20 m vyi = 0

y v yi t  12 a t 2  t 

vx = 2.5 a = g = -10 t = ? x = ?

x v x t  x 2.5 ms (2.0s ) 5.0 m

6. Frustrated with HISTORY, (you never get frustrated in physics) you open the second story classroom window and (to the horror of your teacher but to the secret delight of your classmates) violently hurl your history book out the window with a velocity of 18 m/s at an angle of 35 degrees above the horizontal. If the launch point is 6 meters above the ground, how far from the building will the book hit the parking lot?

v = 18 y

vx = 14.7

m s

35∞

vx

x y = -6.0 m vyi = 10.3 ms

vyi

vx cos35 (18 sm) 14.7 sm vyi  sin35 (18 ms ) 10.3 ms

m

a = g = -10 s 2 t = ? x = ?

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U6 2D Motion - review v3.1

 y v yi t  12 a t 2  -6.0m=(10.3 ms )t 12 ( 10 sm2 )t 2 (  5 ms2 ) t 2  (10.3 ms )t  6.0m 0 (use solver or quadratic) t 2.53 s  x v x  t   x  14.7 ms (2.53s )  37m

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U6 2D Motion - review v3.1...