Question

2-4 The launch velocity of a cannon at A is V-400 m/s. Find the angles θ that will cause the cannon ball to hit the wall at B. (Two values) 1.5 km V-400 m/si 5 km

Answer 1

X = 5km times 2 = 10km = 10000m h = 1.5km = 1500km the horizontal range is defined as X = v^2 sin 2 theta/g = 10000m maximum height y = v^2 sin^2 theta/2g = 1500m dividing X by y v^2 sin^2 theta/2g times g/v^2 sin 2 theta = 1500/10000 sin^2 theta/4 sin theta cos theta = 1500/10000 tan theta = 0.6 theta = tan^-1 (0.6) theta = 31 degree and theta = 149 degree