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Unformatted text preview: 2.41. CHAPTER 2, PROBLEM 41 121 2.41 Chapter 2, Problem 41 Problem: A boy riding in a small cart throws a ball with initial speed vo as seen by a stationary observer at an angle α to the horizontal and catches the ball at a time t after he throws it. The cart follows a straight-line path inclined at an angle φ to the horizontal. The speed of the cart at the instant the boy throws the ball is V and it has constant deceleration of magnitude a. Determine the time t. Ignore effects of friction on the ball’s motion and express your answer as a function of φ, α, vo and gravitational acceleration, g . HINT: Solve using the xz coordinate system and simplify your answer by making use of the fact that sin(α − φ) = sin α cos φ − cos α sin φ. z • vo • α V g = −g k ....... ... ........................................ ................. . ............................................................................................................................................................................................................................. ................................................................................................................................................................................................................................ ...................................................................................................................................................................................................................................... .. ... 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....................................................................................................................................................................................................................................................................................................... ......... ... ... . ................................................................................................................................................................................................................................................................................................................ . φ x So ut on: Ignor ng fr c ona forces he on y force ac ng on he ba s grav y Hence he hor zon a (x) componen of s ve oc y s cons an wh e he ver ca (z ) componen decreases due o g No ng ha we are g ven he abso u e ve oc y e he ve oc y re a ve o a s a onary frame he ve oc y componen s of he ba n a f xed frame of reference are vx = vo cos α and vz = vo s n α − gt In egra ng over me and e ng he ba ’s n a pos on n he f xed coord na e be xB (0) = xo zB (0) = zo he ba ’s pos on n he f xed frame s xB (t) = xo + vo t cos α and zB (t) = zo + vo t s n α − 1 gt2 2 In egra ng and e ng he or g n be he same as he po n where he ba beg ns s ra ec ory he pos on of he car s g ven by xA (t) = xo + V t cos φ − 1 at2 cos φ 2 In genera he re a ve pos on of he ba and zA (t) = zo + V t s n φ − 1 at2 s n φ 2 Turn ng o he car ’s pos on because has cons an dece era on of magn ude a c ear y he ve oc y componen s are vx = V cos φ − at cos φ ˜ and vz = V s n φ − at s n φ ˜ n he car -f xed coord na e sys em s rB/A = rB − rA For he prob em a hand we are seek ng he so u on for wh ch rB/A = 0 Hence we conc ude ha for he x d rec on xo + vo t cos α = xo + V t cos φ − 1 at2 cos φ 2 ⇒ vo cos α = V cos φ 1 − at 2V 122 Similarly, for the z -direction: zo + vo t sin α − 1 gt2 = zo + V t sin φ − 1 at2 sin φ 2 2 Now, from the x-direction equation, there follows 1− vo cos α at = 2V V cos φ ⇒ CHAPTER 2. PARTICLE KINEMATICS vo sin α − 1 gt = V sin φ 1 − 2 at 2V Substituting this result into the z -direction equation yields vo sin α − 1 gt = vo cos α tan φ 2 Hence, we find that 1 2 gt = vo (sin α − cos α tan φ) = vo (sin α cos φ − cos α sin φ) cos φ =sin(α−φ) Therefore, the time, t, is t= 2vo sin(α − φ) g cos φ ...
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