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Unformatted text preview: Homework set 2 Solutions Ch 3 4 • A baseball is hit so its initial velocity upon leaving the bat makes an angle of 30 ° above the horizontal. It leaves that bat at a height of 1.0 m above the ground and lands untouched for a single. During its flight, from just after it leaves the bat to just before it hits the ground, describe how the angle between its velocity and acceleration vectors changes. Neglect any effects due to air resistance. Determine the Concept The angle between its velocity and acceleration vectors starts at 30 ° + 90 ° or 120 ° because the acceleration of the ball is straight down. At the peak of the flight of the ball the angle reduces to 90 ° because the ball’s velocity vector is horizontal. When the ball reaches the same elevation that it started from the angle is 90 ° − 30 ° or 60 ° . 28 •• Two cannons are pointed directly toward each other as shown in Figure 332. When fired, the cannonballs will follow the trajectories shown—P is the point where the trajectories cross each other. If we want the cannonballs to hit each other, should the gun crews fire cannon A first, cannon B first, or should they fire simultaneously? Ignore any effects due to air resistance. Determine the Concept We’ll assume that the cannons are identical and use a constantacceleration equation to express the displacement of each cannonball as a function of time. Having done so, we can then establish the condition under which they will have the same vertical position at a given time and, hence, collide. The modified diagram shown below shows the displacements of both cannonballs. Express the displacement of the cannonball from cannon A at any time t after being fired and before any collision: 52 •• A particle has a constant acceleration of r a = (6.0 m/s 2 ) ˆ i + (4 .0 m/s 2 ) . At time t = 0, the velocity is zero and the position vector is ˆ j r r = (10 m) . ( a ) Find the velocity and position vectors as functions of time t . ˆ i ( b ) Find the equation of the particle’s path in the xy plane and sketch the path. Picture the Problem We can use constantacceleration equations in vector form to find the velocity and position vectors as functions of time t . In ( b ), we can eliminate t from the equations giving the x and y components of the particle to find an expression for y as a function of x . ( a ) Use to find with = + = v t a v v r r r r : v r ( ) ( ) [ ] t j i v ˆ m/s . 4 ˆ m/s . 6 2 2 + = r Use 2 2 1 t t a v r r r r r r + + = ( ) i ˆ m 10 = r : with r to find r r ( ) ( ) [ ] ( [ ] ) j t i t r ˆ m/s . 2 ˆ m/s . 3 m 10 2 2 2 2 + + = r ( b ) Obtain the x and y components of the path from the vector equation in ( a ): ( ) 2 2 m/s . 3 m 10 t x + = and ( ) 2 2 m/s . 2 t y = Eliminate t from these equations and solve for y to obtain: m 3 20 3 2 − = x y The graph of m 3 20 3 2 − = x y is shown below. Note that the path in the xy plane is straight line....
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This note was uploaded on 02/03/2009 for the course PHYS 120 taught by Professor Buck during the Spring '09 term at University of Washington.
 Spring '09
 BUCK
 Physics, Work, Light

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