A cannon fires a 0.652 kg shell with initial
velocity vi = 12 m/s in the direction = 61 ◦
above the horizontal.
The shell’s trajectory curves downward be-
cause of gravity, so at the time t = 0.473 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/s2 .
Answer in units of m
Mass have no effect for the projectile motion and u want to know the height "h" first, find the vertical and horizontal components of velocity vertical component of velocity = 12 sin 61 horizontal component of velocity = 12 cos 61 now for the vertical motion ; S = ut + (1/2) at^2 where s = h u = initial vertical component of velocity t = 0.473 s a = gravitational deceleration (-g) = -9.8 m/s^2
