Brad M. answered • 04/30/13

Calculus for AP, Engineering, and Business VT MATH 1225-6, 1524-6

Hi Ted -- Good to hear from someone in Virginia!

This is a projectile problem that you can sense physically:

Initial velocity launched up at 4 m/s and / 60 degrees, with horizontal component constant x'(0)= 2 m/s and

vertical part y'(0)= 3.7 m/s fighting a constant negative acceleration 10 m/s^2.

After one second, y'(1) suffers a 10 m/s reversal from the starting upward 3.7 ==> -6.3 m/s (downward). Horizontal x'(1) speed remains constant 2 m/s, unchanged by gravity or air drag (assumed).

After two seconds, vertical y"(2) is always -10 m/s/s -- gravity's constant "pull". There can be no horizontal speed change ==> no acceleration ever, x"=0. Warmest regards, sir ...