ch04-p122 - 122. We make use of Eq. 4-21 and Eq.4-22. (a)...

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(c) Differentiating r with respect to t , we obtain 22 2 00 0 2 0 3 sin /2 /4 vv g t g t dr dt g t g t θ −+ = Setting /0 dr dt = , with 0 16.0 m/s v = and 0 40.0 , we have 2 256 151 48 0 tt −+= . The equation has no real solution. This means that the maximum is reached at the end of the flight, with 2 2 sin / 2(16.0 m/s)sin(40.0 )/(9.80 m/s ) 2.10 s. total tv g == ° = 122. We make use of Eq. 4-21 and Eq.4-22. (a) With v o = 16 m/s, we square Eq. 4-21 and Eq. 4-22 and add them, then (using Pythagoras’ theorem) take the square root to obtain r : 2 2 2 0 0 0 0 2 0 () ()( c o s ) ( s i n / 2 ) rx x y y v t g t tv vg t g t θθ =−+ = + =− + Below we plot r as a function of time for o = 40.0º: (b) For this next graph for r versus t we set o = 80.0º.
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(e) The horizontal distance is 00 cos (16.0 m/s)cos40.0 (2.10 s) 25.7 m. x rv t θ == ° = (f) The vertical distance is 0 y r = . (g) For the 0 = 80º launch, the condition for maximum
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ch04-p122 - 122. We make use of Eq. 4-21 and Eq.4-22. (a)...

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