Answers

1. Answer:

   Step-by-step 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 get

   0 = u – 9.8 × 3

   or, u = 29.4 ms-¹

   Thus, the velocity with which it was thrown up = 29.4ms-¹

   ◆ b) Using, 2aS = v² – u², we get

   S = v²- u²/2a

   = 0 – 29.4 × 29.4/- 2× 9.8

   = 44.1 m

   Thus, 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 m

   Therefore, The ball will be 4.9 m below the top of the tower after 4 s

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