EAS208_Exam01_MakeUp_100810_FINAL_Solutions

EAS208_Exam01_MakeUp_100810_FINAL_Solutions - EAS208 Fall...

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EAS208 Fall 2010 – Exam 1 Make-up Exam SOLUTIONS Name: Person Number: Problem 1: The volleyball player serves the ball from point with the speed A 42 o ft v s = at the angle . 28 o θ = a) Derive the equation of the trajectory of the ball ( as a function of y x ) b) Determine whether the ball clears the top of the net C and lands inside the baseline B . Neglect air-resistance. (40 points) Solution: The initial components the ball’s velocity in the x - y coordinate system are: cos xo vv = sin yo = For the motion in the x -direction: 0 x = && 1 0 x xd t d t C = == & 11 2 x t C d t Ct C = + & Invoke initial conditions: () 0c xv C C v o s ==⇒= & ( ) 22 00 xC C 0 = =⇒= For the motion in the -direction: y y g = − ( ) 3 y y dt g dt gt C = + & 2 33 1 2 y y dt gt C dt gt C t C + = + + ∫∫ & 4 i n Invoke initial conditions: 0s yv C C v ==⇒ = & ( ) 44 0 yC C =
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Thus, summarizing for the motion in the x and y direction: () 0 xt = && cos o v θ = & ( ) cos o vt = yt g =− sin o g t v =− + & 2 1 sin 2 o g t vt + a) To find the equation of the trajectory of the ball ( ) yx , solve the equation for x t for time t and substitute in the equation for ( ) : ( ) cos cos o o x t t v =⇒ = thus 2 1 sin 2 o g + 2 22 tan 2c o s o g x x v ⎛⎞ + ⎜⎟ ⎝⎠ () ( ) ( ) 2 0.0117 0.5317 x x + b) The ball is at the net position at C 22 x ft = . From the equation for the trajectory of the ball for 22 x ft = : ( ) 22 6.035 yf t f = t The coordinate system is 3 ft from the ground, thus the ball at the net position is at a height : net ball H ( ) 22 3 9.035 net ball H y ft ft ft =+ = The height of the net is . Therefore, the ball clears the net since:
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EAS208_Exam01_MakeUp_100810_FINAL_Solutions - EAS208 Fall...

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