Question

A ball thrown vertically up returns to the thrower after 6s. Find:

i) The velocity with which it was thrown up.
ii) The maximum height it reaches.
iii) It's position after 4s.​

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

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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Answer 2

Answer:

solved

Explanation:

i) ut - 1/2 t² = 0 , t = 2u/g = u/5 = 6 , u = 30m/s

ii) max.height H = u²/2g = 900/20 = 45m.

iii) its position after 4sec is y = 30x4 - 5x16 = 40m.