Saved Bookmarks
| 1. |
A ball thrown up vertically returns to the thrower after 6sec.Find:- (a) the velocity with which it was thrown up, (b) the maximum height it reaches, and(C) its position after 4 second |
| Answer» EXPLANATION:Time to reach Maximum height ,t = 6/2 = 3 s.v = 0 (at the maximum height )a = - 9.8 m s-²◆ a) Using, v = u + at, we get0 = u - 9.8 × 3or, u = 29.4 ms-¹Thus, the velocity with which it was thrown up = 29.4ms-¹◆ b) Using, 2aS = v² - u², we getS = v²- u²/2a= 0 - 29.4 × 29.4/- 2× 9.8= 44.1 mThus, Maximum height it reaches = 44.1 m.◆ c) Here, t = 4s. In 3 s, the ball reaches the maximum height and in 1 s it falls from the TOP.Distance covered in 1 s from maximum height,S = ut + 1/2at ²= 0 + 1/2 × 9.8 × 1= 4.9 mTherefore, The ball will be 4.9 m below the top of the TOWER after 4 s.hope it helps u plzz mark as BRAINLIEST | |