Question

A ball at point B is at height h above ground level is pushed to the left at an initial velocity v. It reaches point A, which is at a height H above ground level. show that the maximum height of the ball (H) is given by the following equation: H=h+(v^2)/2g

ł ok st =, which is at a height Habove ground leve Aball at point B which is at thight h above ground level is pushed to the eft at an intial velocity v. It reaches poin neo show that the maximum night of the ball (H) is givin by the following equation: H=htu? Zg

Answer 1

4OA H Ih At maximum height, 4 velocity of ball will become zero. Using conservation of energy poinciple- Total mechanical energy at A = Total mechanical energy mgH = mv² + mgh gH v2 + gh at B. H v² zg th