A mechanical launcher placed on the ground can project a ball to a maximum vertical height of 20 m . The same launcher is then placed on the roof of a 45 m tall tower and projects the ball horizontally with the same initial speed. The horizontal distance covered by the ball before hitting the ground is (Take g = 10 m/s ^2 )

Step-by-step solution

The maximum vertical height is achieved when the ball is projected vertically upwards. H_ max = v^2 2g 20 = v^2 2 10 v^2 = 400 v = 20 m/s When the ball is projected horizontally from a height h = 45 m , the time of flight t is given by: t = 2h g = 2 45 10 = 9 = 3 s The horizontal distance covered is: R = v t = 20 3 = 60 m . Answer: 60 m