The initial speed of a cannonball is 260 m/s.
The acceleration of gravity is 9.8 m/s
2
. The
cannon and target are at the same elevation.
If the ball is to strike a target that is at a
horizontal distance of 3.7 km from the cannon,
what is the minimum time of flight for the
ball?
Answer in units of s.
the range is v^2/g sin2θ, so
260^2/9.81 sin2θ = 3700
θ = 16.24°
max height is v^2/2g sin^2θ = 260^2/19.62 * 0.078 = 268.75m
vertical speed v = 260sinθ - 9.81t = 260*0.28 - 9.81t
so it takes 7.42 seconds to reach its max height
Flight time is twice that, or 14.84 seconds