Question

I understand how to do the problem. What I am confused on is more conceptual.

Why is 70.5 the maximum angle? What does it mean when I plug in a smaller angle and get an imaginary answer? Why are larger angles not applicable?

A cannon shoots a cannon ball at a launch angle of theta above the horizontal ground with initial speed v_0. (a) Neglecting air resistance, use Newton's second law to find the ball's position as a function of time. Please use the following coordinates: choose the cannon as the origin, use x to measure horizontal position, and use y to measure vertical position. (b) Let r(t) denote the ball's distance from the cannon. What is the largest possible value of the cannon's launch angle theta if r(t) is to increase throughout the ball's flight?

Answer 1

Imaginary roots for smaller angle theta in the quadratic equation implies that, rate of change of distance, dr/dt , never becomes zero if theta is less than 70.5 deg. That is distance of cannon ball keeps increasing through out it's flight or dr/dt is always +ve.

When ball is launched at angle larger than 70.5 deg, initially dr/dt is +ve, distance increases from the cannon. At some time ( which you can find by plugging actual angle and Vo values in the equation), this dr/dt becomes zero, means ball is at maximum distance from cannon. Thereafter dr/dt becomes -ve and distance starts decreasing. For that = 90 deg, distance keeps increasing till ball is at maximum height, there after distance start decreasing.