Once we know the vertical displacement of the bullet as it passes between the buildings, we can determine the time 1 t required for the bullet to reach the window using Equation 3.4b. Since the motion in the x direction is not accelerated, the distance D can then be found from 0 1 x D v t = . SOLUTION Assuming that the direction to the right is positive, we find that the time that the bullet spends in the building is (according to Equation 3.5a) 0 6.9 m 0.0203 s 340 m/s x x t v = = = The vertical displacement of the bullet after it enters the building is, taking down as the negative direction, equal to –0.50 m. Therefore, the vertical component of the velocity of the bullet as it passes through the window is, from Equation 3.5b, 1 2 1 1 2 2 0 (window) 2 2 –0.50 m ( 9.80 m/s )(0.0203 s) = –24.5 m/s 0.0203 s y y y y a t y v a t t t - = = - = - - The vertical displacement of the bullet as it travels between the buildings is (according to Equation 3.6b with 0 0 y v = m/s) 2 2 2 ( 24.5 m/s) 30.6 m 2 2(–9.80 m/s ) y y v y a - = = = - Therefore, the distance H is 30.6 m 0.50 m 31 m H = + = The time for the bullet to reach the window, according to Equation 3.4b, is
134 KINEMATICS IN TWO DIMENSIONS 1 0 2 2 2(–30.6 m) 2.50 s (–24.5 m/s) y y y y y t v v v = = = = + Hence, the distance D is given by 0 1 (340 m/s)(2.50 s) 850 m x D v t = = = 50. REASONING AND SOLUTION Let H be the initial height of the can above the muzzle of the rifle. Relative to its initial position the vertical coordinate of the bullet at time, t , is 2 1 2 0 y y v t gt = - Relative to the can’s initial position the vertical coordinate of the can at the same time is y ' = (1/2) gt 2 . NOTE: y = H – y ' if the bullet hits the can. Then y = v 0 y t – y '. The horizontal distance traveled by the bullet in time t is x = v 0 x t . Solving for t and substituting gives y = ( v 0 y / v 0 x ) x – y ' Now v 0 y / v 0 x = tan θ and it is seen from the figure in the text that H = x tan θ if the bullet is to hit the can so y = H – y '. Hence, both objects are at the same place at the same time, and the bullet will always strike the can. 51. REASONING The drawing shows the trajectory of the ball, along with its initial speed v 0 , horizontal displacement x , and vertical displacement y . The angle that the initial velocity of the ball makes with the horizontal is θ . The known data are shown in the tables below: x-Direction Data x a x v x v 0 x t +26.9 m 0 m/s 2 +(19.8 m/s) cos θ θ v 0 y x + x + y
Chapter 3 Problems 135 y-Direction Data y a y v y v 0 y t +2.74 m - 9.80 m/s 2 +(19.8 m/s) sin θ There are only two known variables in each table, so we cannot directly use the equations of kinematics to find the angle θ . Our approach will be to first use the x direction data and obtain an expression for the time of flight t in terms of x and v 0 x . We will then enter this expression for t into the y -direction data table. The four variables in this table, y , a y , v 0 y , and t , can be related by using the appropriate equation of kinematics. This equation can then be solved for the angle θ .
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