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A ball thrown up vertically returns …

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A ball thrown up vertically returns to the thrower after 6sec. Find The velocity with which it was thrownup The maximum height it reaches, and Its position after 4sec

Posted by Ishika Awasthi 4 years, 6 months ago

CBSE > Class 09 > Science

1 answers

Yogita Ingle 4 years, 6 months ago

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.

0Thank You

ANSWER

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