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 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