Posted by Noah on .
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
Answer in units of m
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