HW 3 With Solutions

# The vertical component v y of the balls velocity does

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The vertical component v y of the ball’s velocity does change during the flight, since the acceleration in the y direction is that due to gravity ( a y = 9.80 m/s 2 ). The relation (Equation 2.6b) may be used to find , since a y , y , and v 0 y are known ( v 0 y = v 0 sin 40.0º). SOLUTION The speed v of the golf ball just before it lands is v = v x 2 + v y 2 = v 0 cos40.0 ° ( ) 2 + v 0 y 2 + 2 a y y v y 2 = v 0 cos40.0 ° ( ) 2 + v 0 sin40.0 ° ( ) 2 + 2 a y y = 14.0 m/s ( ) cos40.0 ° 2 + 14.0 m/s ( ) sin40.0 ° 2 + 2 9.80 m/s 2 ( ) 3.00 m ( ) = 11.7 m/s 11. Chapter 3, Problem 22 REASONING The vehicle’s initial velocity is in the + y direction, and it accelerates only in the + x direction. Therefore, the y component of its velocity remains constant at v y = +21.0 m/s. Initially, the x component of the vehicle’s velocity is zero ( v 0 x = 0 m/s), but it increases at a rate of a x = +0.320 m/s 2 , reaching a final value v x given by the relation (Equation 3.3a) when the pilot shuts off the RCS

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Chapter 3 4 thruster. Once we have found v x , we will use the right triangle shown in the drawing to find the magnitude v and direction q of the vehicle’s final velocity. SOLUTION a. During the 45.0-second thruster burn, the x component of the vehicle’s velocity increases from zero to Applying the Pythagorean theorem to the right triangle in the drawing, we find the magnitude of the vehicle’s velocity to be: b. Referring again to the drawing, we see that the x and y components of the vehicle’s velocity are related by the tangent of the angle θ . Therefore, the angle of the vehicle’s final velocity is 12. Chapter 3, Problem 24 REASONING In the absence of air resistance, the horizontal velocity component never changes from its initial value of v 0 x . Therefore the horizontal distance D traveled by the criminal (which must equal or exceed the distance between the two buildings) is the initial velocity times the travel time t , or . The time t can be found by noting that the motion of the criminal between the buildings is that of a projectile whose acceleration in the y direction is that due to gravity ( a y = 9.80 m/s 2 , assuming downward to be the negative direction). The relation (Equation 3.5b) allows us to determine the time, since y , v 0 y , and a y are known. Since the criminal is initially running in the horizontal direction, v 0 y = 0 m/s. Setting v 0 y = 0 m/s and solving the equation above for t yields . In this result, y = 2.0 m, since downward is the negative direction. SOLUTION The horizontal distance traveled after launch is . Substituting into this relation gives 13. Chapter 3, Problem 29 REASONING The vertical displacement y of the ball depends on the time that it is in the air before being caught. These variables depend on the y -direction data, as indicated in the table, where the + y direction is "up." Since only two variables in the y direction are known, we cannot determine y at this point. Therefore, we examine the data in the x direction, where + x is taken to be the direction from the pitcher to the catcher.

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