42. A ball is thrown vertically upward from the ground. It crosses a point at the height of 25 m twice at an interval of 4 seconds. The ball was thrown with the velocity of A.18 m/s B.25 m/s C.30 m/s D.36 m/s
Option C Solution: The interval between object pass the same point is 4 sec. That means in 2 sec, object reaches the top and in next 2 sec, it again reaches the same point. By the info given, we can find the velocity of that point by using : v=u+(-gt) 0= u - 10×2 u = 20 Then the velocity at 25m is 20m/sec. so the initial velocity is −v2 = u2 + (−2gh) 400 = u2 − 2x 10 x 25 u2 = 900 => u = 30 m/s
