Question

02. A ball was
thrown Vertically upwards and it came back to
the thrower after 5.5s Find
i) time taken to reach the maximum height.

Answer 1

Answer:

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

0u - 9.8 x 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

the maximum height and in 1 s it falls from the top.

Distance covered in 1 s from maximum

height,

S = ut + 1/2at

= 0 + 1/2 x 9.8 × 1

= 4.9 m

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

Hope it helps

S = v²- u²/2a

= 0 - 29.4 × 29.4/- 2x 9.8

= 44.1 m

Thus, Maximum height it reaches = 44.1 m.

c) Here, t = 4s. In 3 s, the ball reaches