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A cannon fires a 0.639 kg shell with initial
velocity vi = 10 m/s in the direction θ = 55

above the horizontal.
The shell’s trajectory curves downward because of gravity, so at the time t = 0.217 s
the shell is below the straight line by some
vertical distance ∆h.
Find this distance ∆h in the absence of
air resistance. The acceleration of gravity is
9.8 m/s
Answer in units of m

  • physics -

    Mistake: it has to be “above the straight line”

    Δh=vₒ(y) •t-gt²/2= vₒ•sinα•t- gt²/2=
    =10•sin55•0.217 – 9.8•0.217²/2 =
    = 1.78 – 0.23 =1.55 m

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