Posted by beti on .
If the vertical initial speed of the ball is 4.0m/s as the cannon moves horizontally at a speed of 0.70m/s , how far from the launch point does the ball fall back into the cannon?
Vi = 4
h = Hi + Vi t - 4.9 t^2
0 = 0 + 4 t -4.9 t^2
t = 4/4.9 = .816 seconds before it is back down
.816 * . 7 = .571 meters
Your cannonball is remarkably slow, typo?