HW 3 With Solutions

# The five kinematic variables have values one of the

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the five kinematic variables have values, one of the equations of kinematics can be employed to find the acceleration a x . b. The acceleration vector points down and parallel to the ramp, and the angle of the ramp is 25.0 ° relative to the ground (see the drawing). Therefore, trigonometry can be used to determine the component a parallel of the acceleration that is parallel to the ground. SOLUTION a. Equation 3.6a can be used to find the acceleration in terms of the three known variables. Solving this equation for a x gives b. The drawing shows that the acceleration vector is oriented 25.0 ° relative to the ground. The component a parallel of the acceleration that is parallel to the ground is 8. Chapter 3, Problem 12 REASONING AND SOLUTION The vertical motion consists of the ball rising for a time t stopping and returning to the ground in another time t . For the upward portion Note: v y = 0 m/s since the ball stops at the top. Now V 0 y = v 0 sin q 0 = (25.0 m/s) sin 60.0° = 21.7 m/s t = (21.7 m/s)/(9.80 m/s 2 ) = 2.21 s The required "hang time" is 2 t = . 9. Chapter 3, Problem 16 REASONING a. Here is a summary of the data for the skateboarder, using v 0 = 6.6 m/s and q = 58°: y-Direction Data x-Direction Data We will use the relation (Equation 3.6b) to find the skateboarder’s vertical displacement y above the end of the ramp. When the skateboarder is at the highest point, his vertical displacement y 1 above the ground is equal to his initial height y 0 plus his vertical displacement y above the end of the ramp: y 1 = y 0 + y . Next we will use (Equation 3.3b) to calculate the time t from the launch to the highest point, which, with (Equation 3.5a), will give us his horizontal displacement x . SOLUTION x-Direction Data x a x v x v 0 x t +12.0 m ? +7.70 m/s 0 m/s y a y v y v 0 y t ? 9.80 m/s 2 0 m/s +(6.6 m/s)sin 58° = +5.6 m/s x a x v x v 0 x t ? m/s 2 +(6.6 m/s)cos 58° = +3.5 m/s 25.0 ° a parallel a x

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Chapter 3 3 a. Substituting v y = 0 m/s into Equation 3.6b and solving for y , we find Therefore, the highest point y 1 reached by the skateboarder occurs when y 1 = y 0 + y = 1.2 m + 1.6 m = +2.8 m above the ground. b. We now turn to the horizontal displacement, which requires that we first find the elapsed time t . Putting v y = 0 m/s into Equation 3.3b and solving for t yields (1) Given that a x = 0 m/s 2 , the relation (Equation 3.5a) reduces to . Substituting Equation (1) for t then gives the skateboarder’s horizontal displacement: 10. Chapter 3, Problem 20 REASONING The magnitude (or speed) v of the ball’s velocity is related to its x and y velocity components ( v x and v y ) by the Pythagorean theorem; (Equation 1.7). The horizontal component v x of the ball’s velocity never changes during the flight, since, in the absence of air resistance, there is no acceleration in the x direction ( a x = 0 m/s 2 ). Thus, v x is equal to the horizontal component v 0 x of the initial velocity, or . Since v 0 is known, v x can be determined.

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