1)A rock is projected from the edge of the top of a building with an initial vel

Question

1)A rock is projected from the edge of the top of
a building with an initial velocity of 18.6 m/s
at an angle of 32?
above the horizontal. The
rock strikes the ground a horizontal distance
of 98 m from the base of the building.
The acceleration of gravity is 9.8 m/s2.
Assume: The ground is level and that the
side of the building is vertical. Neglect air
friction.What is the horizontal component of the
rock’s velocity when it strikes the ground?
Answer in units of m/s


2)How long does the rock remain airborne?
Answer in units of s

3)How tall is the building? answer in units of m


4)What is the vertical component of the rock’s
velocity when it strikes the ground?
Answer in units of m/s


5)What is the magnitude of the rock’s velocity
when it strikes the ground?
Answer in units of m/s

Explanation / Answer

      The initial velocity of the rock, u = 18.6 m/s       The given angle, = 32o       The horizontal distance, d = 98 m       The acceleration of gravity, g = 9.8 m/s2 _____________________________________________________ 1.       The horizontal component of the
      rock’s velocity when it strikes the ground is                               v0x = ucos                       = (18.6 m/s)cos32o                       = 15.77 m/s _______________________________________________________ _______________________________________________________ 2.       The time taken by the rock in the air is                   t = d/v0x                        = (98 m)/(15.77 m/s)                        = 6.21 s ______________________________________________________ ______________________________________________________ 3.       The initial vertical velocity of the rock is                 v0y = usin                       = (18.6 m/s)sin32o                       = 9.856 m/s                 v0y = usin                       = (18.6 m/s)sin32o                       = 9.856 m/s       Using the kinematic relation, we have              y-y0 = voyt+(1/2)at2               0-y0 = voyt+(1/2)(-g)t2                 -y0 = (9.856 m/s)(6.21 s)-(1/2)(9.8 m/s2)(6.21 s)2       The height of the building is                  y0 = 127.758 m   
                     = 127.6 m    ______________________________________________________ ______________________________________________________ 4.       Using the kinematic relation, we have                    vy = v0y+at                        = (9.856 m/s)+(-9.8 m/s2)(6.21 s)                        = -51.002 m/s                        = (9.856 m/s)+(-9.8 m/s2)(6.21 s)                        = -51.002 m/s       the vertical component of the rock’s
      velocity when it strikes the ground is                    vy= -51.002 m/s       The negative sign indicates the velcity is pointed       to the downward direction.       The negative sign indicates the velcity is pointed       to the downward direction. _______________________________________________________ _______________________________________________________ 5.       The magnitude of the velocity of the rock       when its strike the ground is                  v = (v0x)2+(vy)2                     = (15.77 m/s)2+(-51.002 m/s)2                     = 53.38 m/s                     = 53.38 m/s

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