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A 0.67 kg rock is projected from the edge of
the top of a building with an initial velocity of
11.6 m/s at an angle 52

above the horizontal.
Due to gravity, the rock strikes the ground at
a horizontal distance of 19.5 m from the base
of the building.
How tall is the building? Assume the
ground is level and that the side of the building is vertical. The acceleration of gravity is
9.8 m/s
Answer in units of m

  • physics -

    Upward motion of projectile to the top point
    The height of this point is
    hₒ= vₒ²•sin²α/2g = 11.6²•sin²52/2•9.8 = 4.26 m.
    horizontal displacement is
    L1 = vₒ²•sin2 α/2g=6.6 m.
    Downward motion with horizontal velocity
    v(x) = vₒ•cosα =11.6•cos52=7.14 m/s.
    The horizontal displacement at this motion is
    L2 = L-L1 =19.5-6.66=12.84 m.
    The time of the motion is
    t=L2/v(x) = 12.84/7.14 = 1.79 s.
    The vertical displacement is
    H=gt²/2 =9.8•1.79²/2 = 15.74 m.
    The height of the building is
    h=H-hₒ17.74-4.26 = 11.48 m.

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