v i = 19.0 m/s. h = 41.1 m particle under constant, nonzero acceleration particl

Question

vi = 19.0 m/s. h = 41.1 m particle under constant, nonzero acceleration particle under constant speed    


particle under constant, nonzero acceleration particle under constant speed    





A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of vi = 19.0 m/s. The cliff is h = 41.1 m above a body of water as shown in the figure below.          (a) What are the coordinates of the initial position of the stone?      xi =      m yi =      m     
    (b) What are the components of the initial velocity of the stone?      vix =      m/s viy =      m/s       
    (c) What is the appropriate analysis model for the vertical motion of the stone?
     particle under constant, nonzero acceleration particle under constant speed    



    (d) What is the appropriate analysis model for the horizontal motion of the stone?
     particle under constant, nonzero acceleration particle under constant speed    


  
    (e) Write symbolic equations for the x and y components of the velocity of the stone as a function of time. (Use the following as necessary: vix, g, and t. Indicate the direction of the velocity with the sign of your answer.)      vfx =                          vfy =
                        


    (f) Write symbolic equations for the position of the stone as a function of time. (Use the following as necessary: vix, g, and t. Indicate the direction of the displacement with the sign of your answer.)      xf =                          yf =                                 


    (g) How long after being released does the stone strike the water below the cliff?
          s
    


    (h) With what speed and angle of impact does the stone land?               v =      m/s ? =      A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of vi = 19.0 m/s. The cliff is h = 41.1 m above a body of water as shown in the figure below.

Explanation / Answer

a) Xi = 0 yi = 41.1 m b) vix = 19 m/s viy = 0 c) in x-direction velocity is constant d) in y direction acceleration is constant e) Vfx = vix (Velocity in x-direction does not depend on time) Vfy = vif + g*t f)xf =xi+ Vix*t Yf = yi + viy*t + 0.5*g*t^2 g) t = sqrt(2*s/g) = sqrt(2*41.1/9.8) = 2.986 s h) vfy = g*t = 28.4 m/s v = sqrt(Vfx^2 + Vfy^2) = 34.17 m/s theta = tan^-1(VFy/vFx) = 360-56 = 304 degrees

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