A projectile is shot from the edge of a cliff h = 225 m above ground level with

Question

A projectile is shot from the edge of a cliff h = 225 m above ground level with an initial speed of v0 = 125 m/s at an angle of 37.0° with the horizontal, as shown in Fig. 3-35. (a) Determine the time taken by the projectile to hit point P at ground level.
      s
    (b) Determine the range X of the projectile as measured from the base of the cliff.
      km
    (c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Take up and to the right as positive directions.)
    horizontal m/s
    vertical m/s
    (d) What is the the magnitude of the velocity?
      m/s
    (e) What is the angle made by the velocity vector with the horizontal?
    ° (below the horizontal)
    (f) Find the maximum height above the cliff top reached by the projectile.
      m A projectile is shot from the edge of a cliff h = 225 m above ground level with an initial speed of v0 = 125 m/s at an angle of 37.0° with the horizontal, as shown in Fig. 3-35. (a) Determine the time taken by the projectile to hit point P at ground level.
      s
    (b) Determine the range X of the projectile as measured from the base of the cliff.
      km
    (c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Take up and to the right as positive directions.)
    horizontal m/s
    vertical m/s
    (d) What is the the magnitude of the velocity?
      m/s
    (e) What is the angle made by the velocity vector with the horizontal?
    ° (below the horizontal)
    (f) Find the maximum height above the cliff top reached by the projectile.
      m

Explanation / Answer

Given      height of the cliff   h  = 225 m      initial speed of the projectile    u =   125 m/s      angle of projection    = 37 ^o              vertical inital speed of the projectile at the top of the cliff is   0 m/s      height of the cliff               h = ut + ( 1 /2 ) g t^2                     = gt^2 / 2    1)     time taken by the projectile to reach the ground                t = 2h/g                   =   2 ( 225) / 9.8                   = 6.776 s     2) range X of the projectile as measured from the base of the cliff.
               X = ( u sin ) t                    =( 125 sin 37 )(6.776 )                    = 509.76 m    3) horizontal component of velocity   v_x = u cos                                                              = 125 cos 37                                                              = 99.8294m /s    4) vertical component of velocity    v_y   =   gt   = 9.8 * 6.776 = 66.4048 m /s      5) magnitude of velocity                     v = ( v_x )^2 + ( v_y ) ^2                        = ( 179.6929 )^2 + (66.4048 )^2                           = 119.89 m/s     6) angle made by the velocity vector with the horizontal               tan    = gt / u                         = ( 9.8 ) (6.776) / 125                          = 0.531                       = 27.97 ^o     7) maximum height above the cliff                    H = (125)^2 (sin)^2   / 2 g                        = 125 ^2 ( sin 37 )^2 / ( 2*9.8 )                         = 288.728 m

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